Maths Study Material - Elementary Definite Integral

1. If f & g are functions of x such that g′(x) = f(x) then,

∫ f(x) dx = g(x) + c ⇔ d

dx{g(x)+c} = f(x), where c is called the constant of integration.

2 .2 .2 .2 . Standard Formula:Standard Formula:Standard Formula:Standard Formula:

(i) ∫ (ax + b)n dx =( )

( )ax b

a n

n+

+

+1

1 + c, n ≠ −1

(ii) ∫ dx

ax b+ =

1

a ln (ax + b) + c

(iii) ∫ eax+b dx =1

a eax+b + c

(iv) ∫ apx+q dx =1

p

a

na

px q+

� + c; a > 0

(v) ∫ sin (ax + b) dx = −1

a cos (ax + b) + c

(vi) ∫ cos (ax + b) dx =1

a sin (ax + b) + c

(vii) ∫ tan(ax + b) dx =1

a ln sec (ax + b) + c

(viii) ∫ cot(ax + b) dx =1

a ln sin(ax + b)+ c

(ix) ∫ sec² (ax + b) dx =1

a tan(ax + b) + c

(x) ∫ cosec²(ax + b) dx = −1

acot(ax + b)+ c

(xi) ∫ sec (ax + b). tan (ax + b) dx =1

a sec (ax + b) + c

(xii) ∫ cosec (ax + b). cot (ax + b) dx = −1

a cosec (ax + b) + c

(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π

4 2+

x+ c

(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tanx

2

(xv) ∫d x

a x2 2− = sin−1

x

a + c

(xvi) ∫d x

a x2 2+

=1

a tan−1

x

a + c

(xvii) ∫d x

x x a2 2− =

1

a sec−1

x

a + c

(xviii) ∫d x

x a2 2+

= ln [ ]x x a+ +2 2 OR sinh−1

x

a + c

(xix) ∫d x

x a2 2− = ln [ ]x x a+ −2 2

OR cosh−1x

a + c

(xx) ∫d x

a x2 2−

=1

2a ln

xa

xa

−

+ + c

(xxi) ∫d x

x a2 2− =

1

2a ln

ax

ax

+

− + c

(xxii) ∫ a x2 2− dx =x

2a x2 2− +

a2

2 sin−1

x

a + c

(xxiii) ∫ x a2 2+ dx =x

2x a2 2+ +

a2

2 �n

++

a

axx 22

+ c

(xxiv) ∫ x a2 2− dx =x

2x a2 2− −

a2

2 �n

−+

a

axx 22

+ c

(xxv) ∫ eax. sin bx dx =e

a b

ax

2 2+ (a sin bx − b cos bx) + c

(xxvi) ∫ eax. cos bx dx =e

a b

ax

2 2+ (a cos bx + b sin bx) + c

3 .3 .3 .3 . Theorems on integrat ionTheorems on integrat ionTheorems on integrat ionTheorems on integrat ion

(i) ∫ dx).x(fc = c ∫ dx).x(f

(ii) ∫ ± dx))x(g)x(f( = ∫ ± dx)x(gdx)x(f

(iii) ∫ += c)x(gdx)x(f ⇒ ∫ + dx)bax(f = a

)bax(g + + c

Note : (i) every contineous function is integrable

(ii) the integral of a function reffered only by a constant.

∫ dx).x(f = g(x) + c

= h(x) + c

g′(x) = f(x) & h′(x) = f(x)

g′(x) – h′(x) = 0

means, g(x) – h(x) = c

Example : Evaluate : ∫ dxx4 5

Solution. ∫ dxx4 5 =

6

4 x6 + C =

3

2 x6 + C.

Example : Evaluate : ∫

++−+ dx

x

2

x

74x5x 23

Solution. ∫

++−+ dx

x

2

x

74x5x 23

= ∫ dxx3 + ∫ dxx5 2

– ∫ dx4 + ∫ dxx

7 + ∫ dx

x

2

= ∫ dxx3 + 5 . ∫ dxx2

– 4 . ∫ dx.1 + 7 . ∫ dxx

1 + 2 . ∫

− dxx 2/1

= 4

x4

+ 5 . 3

x3

– 4x + 7 log | x | + 2

2/1

x 2/1

+ C

= 4

x4

+ 3

5x3 – 4x + 7 log | x | + 4 x + C

Example : Evaluate : ∫ ++ alogaxlogaalogx eee dx

Solution. We have,

∫ ++ alogaxlogaalogx eee dx

= ∫ ++aax alogxlogalog eee dx

= ∫ ++ )axa( aax dx

= ∫ dxax + ∫ dxxa

+ ∫ dxaa

= alog

ax

+ 1a

x 1a

+

+

+ aa . x + C.

Example : Evaluate : ∫+x

xx

5

32 dx

Solution. ∫+x

xx

5

32 dx

= ∫

+

x

x

x

x

5

3

5

2 dx

= ∫

+

xx

5

3

5

2 dx =

5/2log

)5/2(

e

x

+ 5/3log

)5/3(

e

x

+ C

Example: Evaluate : ∫ xcosxsin 33 dx

Solution. = 8

1

3)xcosxsin2(∫ dx

= 8

1 ∫ x2sin3 dx

= 8

1 ∫

−

4

x6sinx2sin3 dx

= 32

1 ∫ − )x6sinx2sin3( dx

= 32

1

+− x6cos

6

1x2cos

2

3 + C

Example : Evaluate : ∫ +1x

x2

4

dx

Solution. ∫ +1x

x2

4

dx

= ∫ +

+−

1x

11x2

4

dx = ∫ +

−

1x

1x2

4

+ 1x

12 +

dx

= )1x( 2 −∫ dx + ∫ +1x

12 dx =

3

x3

– x + tan–1 x + C

Example: Evaluate : ∫ + 2x94

1 dx

Solution. We have

∫ + 2x94

1

= 9

1 ∫

+ 2x9

4

1 dx

= 9

1 ∫ + 22 x)3/2(

1 dx

= 9

1 .

)3/2(

1 tan–1

3/2

x + C =

6

1 tan–1

2

x3 + C

Example : ∫ dxx2cosxcos

Solution. ∫ dxx2cosxcos

= 2

1∫ dxx2cosxcos2

= 2

1∫ + )xcosx3(cos dx

= 2

1

+

1

xsin

3

x3sin + c

Self Practice Problems

1. Evaluate : ∫ xtan2 dx Ans. tanx – x + C

2. Evaluate : ∫ + xsin1

1 dx Ans. tanx – sec x + C

4.4 .4 .4 . Integration by SubsitutionsIntegration by SubsitutionsIntegration by SubsitutionsIntegration by Subsitutions

If we subsitute x = φ(t) in a integral then(i) everywhere x will be replaced in terms of t.(ii) dx also gets converted in terms of dt.(iii) φ(t) should be able to take all possible value that x can take.

Example : Evaluate : ∫43 xsinx dx

Solution. We have

Ι = ∫43 xsinx dx

L e t xL e t xL e t xL e t x

4 = t ⇒ d(x4) = dt ⇒ 4x3 dx = dt ⇒ dx = 3x4

1 dt

Example : ∫ x

)xn( 2�

dx

Solution. ∫ x

)xn( 2�

dx

Put �nx = t

⇒x

1 dx = dt

= ∫

x

dx.t2

= ∫ dtt2

= 3

t3

+ c

)xn( 3

Example : Evaluate ∫ + dxxcos)xsin1( 2

Solution. Put sinx = tcosx dx = dt

∫ + dt)t1( 2 = t +

3

t3

+ c

� + c=

3

= sin x + 3

xsin3

+ c

Example : Evaluate : ∫ ++ 1xx

x24 dx

Solution. We have,

Ι = ∫ ++ 1xx

x24 dx = ∫ ++ 1x)x(

x222 dx

Let x2 = t, then, d (x2) = dt ⇒ 2x dx = dt ⇒ dx = x2

dt

Ι = ∫ ++ 1tt

x2 .

x2

dt

= 2

1 ∫ ++ 1tt

12 dt

= 2

1 ∫

+

+

22

2

3

2

1t

1 dt

= 2

1 .

2

3

1 tan–1

+

2

3

2

1t

+ C

= 3

1 tan–1

+

3

1t2 + C =

3

1 tan–1

+

3

1x2 2

+ C.

Note: (i) ∫ [ f(x)]n f ′(x) dx = 1n

))x(f( 1n

+

+

(ii) ∫[ ]

′f x

f xn

( )

( ) dx =

n1

))x(f( n1

−

−

(iii) ∫ d x

x xn

( )+1 n ∈ N Take xn common & put 1 + x−n = t.

(iv) ∫( )

dx

x xnn

n21

1+−( ) n ∈ N, take xn common & put 1+x−n = tn

(v)

( )dx

x xn n

n

11

+∫ / take xn common as x and put 1 + x −n = t.


1. ∫ +dx

xtan1

xsec2

Ans. �n |1 + tan x| + C

2. ∫ dxx

)nxsin(�Ans. – cos (�n x) + C

5 .5 .5 .5 . Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :

( )∫ )x(g)x(f dx = f(x) ( )∫ )x(g dx – ( ) ( ) dxdx)x(g)x(fdx

d∫ ∫

(i) when you find integral ∫ dx)x(g then it will not contain arbitarary constant.

(ii) ∫ dx)x(g should be taken as same both terms.

(iii) the choice of f(x) and g(x) is decided by ILATE rule.

the function will come later is taken an integral function.

Ι → Inverse function

L → Logrithimic function

A → Algeberic function

T → Trigonometric function

E → Exponential function

Example : Evaluate : ∫− dxxtanx 1

Solution. ∫− dxxtanx 1

= (tan–1 x) 2

x2

– ∫ + 2x1

1 .

2

x2

dx

= 2

x2

tan–1 x – 2

1 ∫ +

−+

1x

11x2

2

dx = 2

x2

tan–1 x – 2

1 ∫ +

−1x

11

2 dx

= 2

x2

tan–1 x – 2

1 [x – tan–1 x] + C.

Example : Evaluate : ∫ + dx)x1log(x

Solution. ∫ + dx)x1log(x

= log (x + 1) . 2

x2

– ∫ +1x

1 .

2

x2

dx

= 2

x2

log (x + 1) – 2

1 ∫ +1x

x2

dx = 2

x2

log (x + 1) – 2

1 ∫ +

+−

1x

11x2

dx

= 2

x2

log (x + 1) – 2

1 ∫ +

−

1x

1x2

+ 1x

1

+ dx

= 2

x2

log (x + 1) – 2

1

++−∫ dx

1x

1)1x(

= 2

x2

log (x + 1) – 2

1

++− |1x|logx

2

x2

+ C

Example : Evaluate : ∫ x3sine x2 dx

Solution. Let Ι = ∫ x3sine x2dx. Then,

Ι = ∫ x3sine x2dx

⇒ Ι = e2x

−

3

x3cos – ∫

x2e2

−

3

x3cos dx

⇒ Ι = – 3

1 e2x cos 3x +

3

2 ∫ x3cose x2

dx

⇒ Ι = – 3

1 e2x cos 3x +

3

2

− ∫ dx

3

x3sine2

3

x3sine x2x2

⇒ Ι = – 3

1 e2x cos 3x +

9

2e2x sin 3x –

9

4 ∫ x3sine x2

dx

⇒ Ι = – 3

1 e2x cos 3x +

9

2e2x sin 3x –

9

4 Ι

⇒ Ι + 9

4 Ι =

9

e x2

(2 sin 3x – 3 cos 3x)

⇒9

13 Ι =

9

e x2

(2 sin 3x – 3 cos 3x)

⇒ Ι = 13

e x2

(2 sin 3x – 3 cos 3x) + C

Note :

(i) ∫ ex [f(x) + f ′(x)] dx = ex. f(x) + c

(ii) ∫ [f(x) + xf ′(x)] dx = x f(x) + c

Example : ∫xe 2)1x(

x

+ dx

Solution. ∫xe 2)1x(

11x

+

−+ dx

⇒ ∫ xe

+−

+ 2)1x(

1

)1x(

1 dx =

)1x(

ex

+ + c

Example : ∫ xe

−

−

xcos1

xsin1 dx

Solution. ∫xe

−

2

xsin2

2

xcos

2

xsin21

2 dx

⇒ ∫xe

−

2

xcoteccos

2

1 2 dx = – ex cot

2

x + c

Example : ∫

+

2)nx(

1)nx(n�

�� dx

Solution. put x = et

⇒ ∫te

+

2t

1nt� dt

⇒ ∫te

++−

2t

1

t

1

t

1nt� dt = et

−

t

1nt� + c

⇒ x

−

nx

1)nx(n�

�� + c


1. ∫ dxxsinx Ans. – x cosx + sin x + C

2. ∫ dxex x2Ans. x2 ex – 2xex + 2ex + C

6.6 .6 .6 . Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:

PARTIAL FRACTIONS :

If f(x) and g(x) are two polynomials, then )x(g

)x(f defines a rational algebraic function of a rational function

of x.

If degree of f(x) < degree of g(x), then )x(g

)x(f is called a proper rational function.

If degree of f(x) ≥ degree of g(x) then )x(g

)x(f is called an improper rational function

If )x(g

)x(f is an improper rational function, we divide f(x) by g(x) so that the rational function

)x(g

)x(f is

expressed in the form φ(x) + )x(g

)x(Ψ where φ(x) and ψ(x) are polynomials such that the degree of ψ(x) is

less than that of g(x). Thus, )x(g

)x(f is expressible as the sum of a polynomial and a proper rational

function.

Any proper rational function )x(g

)x(f can be expressed as the sum of rational functions, each having a

simple factor of g(x). Each such fraction is called a partial fraction and the process of obtained them is

called the resolutions or decomposition of )x(g

)x(f into partial fractions.

The resolution of )x(g

)x(f into partial fractions depends mainly upon the nature of the factors of g(x) as

discussed below.

CASE I When denominator is expressible as the product of non-repeating linear factors.

Let g(x) = (x – a1) (x – a

2) .....(x – a

n). Then, we assume that

)x(g

)x(f =

1

1

ax

A

− +

2

2

ax

A

− + ..... +

n

n

ax

A

−

where A1, A

2, ...... A

n are constants and can be determined by equating the numerator on R.H.S. to the

numerator on L.H.S. and then substituting x = a1, a

2, ........,a

n.

Example : Resolve 6x11x6x

2x323 −+−

+ into partial fractions.

Solution. We have, 6x11x6x

2x323 −+−

+ =

)3x)(2x)(1x(

2x3

−−−

+

Let)3x)(2x)(1x(

2x3

−−−

+ =

1x

A

− +

2x

B

− +

3x

B

−. Then,

⇒)3x)(2x)(1x(

2x3

−−−

+ =

)3x)(2x)(1x(

)2x)(1x(C)3x)(1x(B)3x)(2x(A

−−−

−−+−−+−−

⇒ 3x + 2 = A(x – 2) (x – 3) + B (x – 1) (x – 3) + C(x – 1) (x – 2) ...........(i)

Putting x – 1 = 0 or x = 1 in (i), we get

5 = A(1 – 2) (1 – 3) ⇒ A = 2

5,

Putting x – 2 = 0 or, x = 2 in (i), we obtain

8 = B (2 – 1) (2 – 3) ⇒ B = –8.

Putting x – 3 = 0 or, x = 3 in (i), we obtain

11 = C (3 – 1) (3 – 2) ⇒ C = 2

11.

∴6x11x6x

2x323 −+−

+ =

)3x)(2x)(1x(

2x3

−−−

+ =

)1x(2

5

− –

2x

8

− +

)3x(2

11

−

Note : In order to determine the value of constants in the numerator of the partial fraction corresponding to the

non-repeated linear factor px + q in the denominator of a rational expression, we may proceed as

follows :

Replace x = – p

q (obtained by putting px + q = 0) everywhere in the given rational expression except in

the factor px + q itself. For example, in the above illustration the value of A is obtained by replacing x

by 1 in all factors of )3x)(2x)(1x(

2x3

−−−

+ except (x – 1) i.e.

A = )31)(21(

213

−−

+× =

2

5

Similarly, we have

B = )32)(21(

123

−−

+× = –8 and, C =

)23)(13(

233

−−

+× =

2

11

Example : Resolve 6x5x

2x10x6x2

23

+−

−+− into partial fractions.

Solution. Here the given function is an improper rational function. On dividing we get

6x5x

2x10x6x2

23

+−

−+− = x – 1 +

)6x5x(

)4x(2 +−

+−...........(i)

we have, 6x5x

4x2 +−

+− =

)3x)(2x(

4x

−−

+−

So, let )3x)(2x(

4x

−−

+− =

2x

A

− +

3x

B

− – x + 4 = A(x – 3) + B(x – 2) ...........(ii)

Putting x – 3 = 0 or, x = 3 in (ii), we get

1 = B(1) ⇒ B = 1.

Putting x – 2 = 0 or, x = 2 in (ii), we get

2 = A (2 – 3) ⇒ A = – 2

∴)3x)(2x(

4x

−−

+− =

2x

2

−

− +

3x

1

−

Hence6x5x

2x10x6x2

23

+−

−+− = x – 1 –

2x

2

− +

3x

2

−

CASE II When the denominator g(x) is expressible as the product of the linear factors such that some

of them are repeating.

Example )x(g

1 =

)ax).......(ax)(ax()ax(

1

r21k −−−−

this can be expressed as

ax

A1

− + 2

2

)ax(

A

− + 3

3

)ax(

A

− + ....+ k

k

)ax(

A

− +

)ax(

B

1

1

− +

)ax(

B

2

2

− + ...... +

)ax(

B

r

r

−

Now to determine constants we equate numerators on both sides. Some of the constants are determined

by substitution as in case I and remaining are obtained by

The following example illustrate the procedure.

Example : Resolve )2x)(1x()1x(

2x32 ++−

− into partial fractions, and evaluate ∫ ++−

−

)2x)(1x()1x(

dx)2x3(2

Solution. Let )2x)(1x()1x(

2x32 ++−

− =

1x

A1

− + 2

2

)1x(

A

− +

1x

A3

+ +

2x

A4

+

⇒ 3x – 2 = A1 (x – 1) (x + 1) (x + 2) + A

2 (x + 1) (x + 2)

+ A3 (x – 1)2 (x + 2) + A

4 (x – 1)2 (x + 1) .......(i)

Putting x – 1 = 0 or, x = 1 in (i) we get

1 = A2 (1 + 1) (1 + 2) ⇒ A

2 =

6

1

Putting x + 1 = 0 or, x = –1 in (i) we get

– 5 = A3 (–2)2 (–1 + 2) ⇒ A

3 = –

4

5

Putting x + 2 = 0 or, x = –2 in (i) we get

– 8 = A4 (–3)2 (–1) ⇒ A

4 =

9

8

Now equating coefficient of x3 on both sides, we get 0 = A1 + A

3 + A

4

⇒ A1 = –A

3 – A

4 =

4

5 –

9

8 =

36

13

∴)2x)(1x()1x(

2x32 ++−

− =

)1x(36

13

− + 2)1x(6

1

− –

)1x(4

5

+ +

)2x(9

8

+

and hence ∫ ++−

−

)2x)(1x()1x(

dx)2x3(2

=36

13 �n |x – 1| –

)1x(6

1

− –

4

5 �n |x + 1| +

9

8 �n |x + 2| + c

CASE III When some of the factors of denominator g(x) are quadratic but non-repeating. Corresponding

to each quadratic factor ax2 + bx + c, we assume partial fraction of the type cbxax

BAx2 ++

+, where A and

B are constants to be determined by comparing coefficients of similar powers of x in the numerator of

both sides. In practice it is advisable to assume partial fractions of the type cbxax

)bax2(A2 ++

+ +

cbxax

B2 ++

The following example illustrates the procedure

Example : Resolve )2x)(1x(

1x22 ++

− into partial fractions and evaluate ∫ ++

−

)2x)(1x(

1x22 dx

Solution. Let)2x)(1x(

1x22 ++

− =

1x

A

+ +

2x

CBx2 +

+. Then,

)2x)(1x(

1x22 ++

− =

)2x)(1x(

)1x)(CBx()2x(A2

2

++

++++

⇒ 2x – 1 = A (x2 + 2) + (Bx + C) (x + 1) ...(i)

Putting x + 1 = 0 or, x = –1 in (i), we get – 3 = A(3) ⇒ A = –1.

Comparing coefficients of the like powers of x on both sides of (i), we get

A + B = 0, C + 2A = –1 and C + B = 2

∴ –1 + B = 0, C – 2 = –1 (Putting A = –1)

⇒ B = 1, C = 1

∴)2x)(1x(

1x22 ++

− = –

1x

1

+ +

2x

1x2 +

+

Hence ∫ ++

−

)2x)(1x(

1x22 dx

= – �n |x + 1| + 2

1 �n |x2 + 1| +

2

1 tan–1

2

x + c

CASE IV When some of the factors of the denominator g(x) are quadratic and repeating fractions of the

form

+++

++

+

cbxax

A

cbxax

)bax2(A2

1

2

0 + ( ) ( )

+++

++

+22

2

22

1

cbxax

A

cbxax

)bax2(A

+ .......+ ( ) ( )

+++

++

+−

k2

k2

k2

1k2

cbxax

A

cbxax

)bax2(A

The following example illustrates the procedure.

Example: Resolve 22 )1x)(1x(

3x2

+−

− into partial fractions.

Solution. Let 22 )1x)(1x(

3x2

+−

− =

1x

A

− +

1x

CBx2 +

+ + 22 )1x(

EDx

+

+. Then,

2x – 3 = A(x2 + 1)2 + (Bx + C) (x – 1) (x2 + 1) + (Dx + E) (x – 1) ......(i)

Putting x = 1 in (i), we get – 1 = A (1 + 1)2 ⇒ A = – 4

1

Equation coefficients of like powers of x, we have

A + B = 0, C – B = 0, 2A + B – C + D = 0, C + E – B – D = 2 and A – C – E = –3.

Putting A = – 4

1 and solving these equations, we get

B = 4

1 = C, D =

2

1 and E =

2

5

∴ 22 )1x)(1x(

3x2

+−

− =

)1x(4

1

−

− +

)1x(4

1x2 +

+ + 22 )1x(2

5x

+

+

Example : Resolve 1x

x23 −

into partial fractions.

Solution. We have, 1x

x23 −

= )1xx)(1x(

x22 ++−

So, let )1xx)(1x(

x22 ++−

= 1x

A

− +

1xx

CBx2 ++

+. Then,

2x = A (x2 + x + 1) + (Bx + C) (x – 1) .......(i)

Putting x – 1 = 0 or, x = 1 in (i), we get 2 = 3 A ⇒ A = 3

2

Putting x = 0 in (i), we get A – C = 0 ⇒ C = A = 3

2

Putting x = – 1 in (i), we get –2 = A + 2B – 2 C.

⇒ – 2 = 3

2 + 2B –

3

4 ⇒ B = –

3

2

∴1x

x23 −

= 3

2 .

1x

1

− +

1xx

3/2x3/22 ++

+− or, ,

1x

x23 −

= 3

2

1x

1

− +

3

2

1xx

x12 ++

−


1. (i) ∫ ++dx

)3x)(2x(

1Ans. �n

3x

2x

+

+ + C

(ii) ∫ ++ )1x)(1x(

dx2 Ans.

2

1 �n |x + 1| –

4

1 �n (x2 + 1) +

2

1 tan–1 (x) + C2222 , ∫

++ ccccbxbxbxbxaxaxaxax dxdxdxdx2222 , ∫ ++ ccccbxbxbxbxaxaxaxax 2222 dx

Express ax2 + bx + c in the form of perfect square & then apply the standard results.

Example : Evaluate : ∫ ++ 5x2x2 dx

Solution. We have,

2

2 dx= ∫ x + 2x + 1+ 4

∫ x + 2x + 5

7777 .... IIIInnnntttteeeeggggrrrraaaatttt iiiioooonnnn ooooffff ttttyyyyppppeeee ∫ axaxaxax +bxbxbxbx+ccccdxdxdxdx

= 2

1 (x + 1) 22 2)1x( +− +

2

1 . (2)2 log |(x + 1) + 22 2)1x( ++ | + C

= 2

1 (x + 1) 5x2x2 +− + 2 log |(x + 1) + 5x2x2 ++ | + C

Example : Evaluate : ∫ +− 1xx

12 dx

Solution. ∫ +− 1xx

12 dx

= ∫+−+− 1

4

1

4

1xx

1

2 dx

= ∫ +− 4/3)2/1x(

12 dx

= ( )∫+−

22 2/3)2/1x(

1

dx = 2/3

1 tan–1

−

2/3

2/1x + C

= 3

2 tan–1

−

3

1x2 + C.

Example : Evaluate : ∫−+ 2xx89

1 dx

Solution. ∫−+ 2xx89

1 dx

= ∫−−− }9x8x{

1

2 dx

= ∫−+−− }2516x8x{

1

2 dx

= ∫ −−− }5)4x{(

122 dx = ∫

−− 22 )4x(5

1 dx = sin–1

−

5

4x + C


1. ∫ −+ 1xx2

12 dx Ans.

3

1 �n

2x2

1x2

+

− + C

2. ∫−+ 2x3x2

1

2 dx Ans.2

1 log 1x

2

3x

4

3x 2 −++

+ + C

8 .8 .8 .8 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type

∫ ++

+ ccccbxbxbxbxaxaxaxax qqqqpxpxpxpx2222 dx, ∫++

+ ccccbxbxbxbxaxaxaxax qqqqpxpxpxpx2222 dx, ∫∫∫∫ ++++++++++++ cbxax)qpx( 2 dx

Express px + q = A (differential co−efficient of denominator) + B.

Example : Evaluate : ∫++

+

1x4x

3x2

2 dx

Solution. ∫++

+

1x4x

3x2

2 dx

= ∫++

−+

1x4x

1)4x2(

2 dx

= ∫++

+

1x4x

4x2

2 dx – ∫++ 1x4x

1

2 dx

= ∫ t

dt –

( )∫

−+22 3)2x(

1 dx, where t = x2 + 4x + 1

= 2 t – log | (x + 2) + 1x4x2 ++ | + C

= 2 1x4x2 ++ – log | x + 2 + 1x4x2 ++ | + C

Example : Evaluate : ∫ +− xx)5x( 2 dx

Solution. Let (x – 5) = λ . dx

d (x2 + x) + µ. Then,

x – 5 = λ (2x + 1) + µ.

Comparing coefficients of like powers of x, we get

1 = 2λ and λ + µ = – 5 ⇒ λ = 2

1 and µ = –

2

11

∫ +− xx)5x( 2 dx

= ∫

−+

2

11)1x2(

2

1 xx2 + dx

= ∫ + )1x2(2

1 xx2 + dx –

2

11 ∫ + xx2

dx

= 2

1∫ + )1x2( xx2 + dx –

2

11 ∫ + xx2

dx

= 2

1

∫ t dt – 2

11 ∫

−

+

22

2

1

2

1x dx where t = x2 + x

= 2

1 .

2/3

t 2/3

– 2

11

−

+

+

22

2

1

2

1x

2

1x

2

1

– 2

1 .

2

2

1

log

−

++

+

22

2

1

2

1x

2

1x + C

= 3

1 t3/2 –

2

11

++

+−+

+xx

2

1xn

8

1xx

4

1x2 22� + C

= 3

1(x2 + x)3/2 –

2

11

++

+−+

+xx

2

1xn

8

1xx

4

1x2 22� + C


1. ∫ ++

+

3xx

1x2 dx Ans.

2

1 log |x2 + x + 3| +

11

1 tan–1

+

11

1x2 + C

2. ∫+−

−

1x5x3

5x6

2 dx Ans. 2 1x5x3 2 +− + C

3. ∫ ++− 2xx1)1x( dx

Ans.3

1 (x2 + x + 1)3/2 –

8

3 (2x + 1) 2xx1 ++ –

16

9 log (2x +1 + 2 1xx2 ++ ) + C

9.9 .9 .9 . Integrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functions

(i) ∫ xxxxsinsinsinsinbbbbaaaa xxxxdddd 2222+ OR ∫ xxxxcoscoscoscosbbbbaaaa xxxxdddd 2222+

OR ∫ xxxxcoscoscoscosccccxxxxcoscoscoscosxxxxsinsinsinsinbbbbxxxxsinsinsinsinaaaa xxxxdddd 22222222 ++

Multiply Nr & Dr by sec² x & put tan x = t.

(ii) ∫ sinxsinxsinxsinxbbbbaaaa xxxxdddd+

OR ∫ cosxcosxcosxcosxbbbbaaaa xxxxdddd+

OR ∫ xxxxcoscoscoscosccccxxxxsinsinsinsinbbbbaaaa xxxxdddd++

Hint:

Convert sines & cosines into their respective tangents of half the angles and then,

put tan2

x = t

(iii) ∫ nnnnxxxxsinsinsinsin....mmmmxxxxcoscoscoscos.... ccccxxxxsinsinsinsin....bbbbxxxxcoscoscoscos....aaaa++

++

� dx. Express Nr ≡≡≡≡ A(Dr) + B

d

d x (Dr) + c & proceed.

Example : Evaluate : ∫ ++ xcosxsin1

1 dx

Solution. Ι = ∫ ++ xcosxsin1

1 dx

= ∫

+

−+

++

2/xtan1

2/xtan1

2/xtan1

2/xtan21

1

2

2

2 dx

= ∫ −+++

+

2/xtan12/xtan22/xtan1

2/xtan122

2

dx = ∫ + 2/xtan22

2/xsec2

dx

Putting tan2

x = t and

2

1 sec2

2

x dx = dt, we get

Ι = ∫ + 1t

1 dt = log | t + 1| + C = log 1

2

xtan + + C

Example : Evaluate : ∫ +

+

xsin2xcos3

xcos2xsin3 dx

Solution. Ι = ∫ +

+

xsin2xcos3

xcos2xsin3 dx

Let 3 sin x + 2 cos x = λ. dx

d (3 cos x + 2 sin x ) + µ (3 cos x + 2 sin x)

⇒ 3 sin x + 2 cos x = λ (–3 sin x + 2 cos x) + µ (3 cos x + 2 sin x )

Comparing the coefficients of sin x and cos x on both sides, we get

– 3λ + 2µ = 3 and 2λ + 3µ = 2 ⇒ µ = 13

12 and λ = –

13

5

∴ Ι = ∫ +

+λ++−µ

xsin2xcos3

)xsin2xcos3()xcos2xsin3( dx

= λ ∫ dx.1 + µ ∫ +

+−

xsin2xcos3

xcos2xsin3 dx

= λ x + µ ∫ t

dt, where t = 3 cos x + 2 sin x

= λ x + µ �n | t | + C = 13

5− x +

13

12 �n | 3 cos x + 2 sin x | + C

Example : Evaluate : ∫ ++

+

3xcos2xsin

2xcos3 dx

Solution. We have,

Ι = ∫ ++

+

3xcos2xsin

2xcos3 dx

Let 3 cos x + 2 = λ (sin x + 2 cos x + 3) + µ (cos x – 2 sin x) + ν

Comparing the coefficients of sin x, cos x and constant term on both sides, we get

λ – 2µ = 0, 2λ + µ = 3, 3λ + ν = 2

⇒ λ = 5

6, µ

5

3 and ν = –

5

8

∴ Ι = ∫ ++

ν+−µ+++λ

3xcos2xsin

)xsin2x(cos)3xcos2x(sin dx

⇒ Ι = λ ∫ µ+dx ∫ ++

−

3xcos2xsin

xsin2xcos dx + ν ∫ ++ 3xcos2xsin

1 dx

⇒ Ι = λ x + µ log | sin x + 2 cos x + 3 | + ν Ι1, where

Ι1 = ∫ ++ 3xcos2xsin

1 dx

Putting, sin x = 2/xtan1

2/xtan22+

, cos x = 2/xtan1

2/xtan12

2

+

− we get

Ι1 = ∫

++

−+

+3

2/xtan1

)2/xtan1(2

2/xtan1

2/xtan2

1

2

2

2

dx

= ∫ ++−+

+

)2/xtan1(32/xtan222/xtan2

2/xtan122

2

dx

= ∫ ++ 52/xtan22/xtan

2/xsec2

2

dx

Putting tan 2

x = t and

2

1 sec2

2

x = dt or sec2

2

x dx = 2 dt, we get

Ι1 = ∫ ++ 5t2t

dt22

= 2 ∫ ++ 22 2)1t(

dt =

2

2 tan–1

+

2

1t = tan–1

+

2

12

xtan

Hence, Ι = λx + µ log | sin x + 2 cos x + 3 | + ν tan–1

+

2

12

xtan

+ C

where λ = 5

6, µ =

5

3 and ν = –

5

8

Example : ∫ + xcos31

dx2

Solution. = ∫ + 4xtan

dxxsec2

2

= 2

1 tan–1

2

xtan + C


1. ∫ +

+

xcos4xsin5

xcos5xsin4 dx Ans.

41

40x +

41

9 log |5sinx + 4cosx| + C

10 .10 .10 .10 . Integration of type Integration of type Integration of type Integration of type ∫ dxdxdxdxxxxxcoscoscoscosx.x.x.x.sinsinsinsin nnnnmmmmCase - ΙΙΙΙ

If m and n are even natural number then converts higher power into higher angles.

Case - ΙΙΙΙΙΙΙΙ

If at least m or n is odd natural number then if m is odd put cosx = t and vice-versa.

Case - ΙΙΙ ΙΙΙ ΙΙΙ ΙΙΙ

When m + n is a negative even integer then put tan x = t.

Example: ∫ dxxcosxsin 45

Solution. put cos x = t ⇒ – sinx dx = dt

= – ∫ − 22 )t1( . t4 . dt

= – ∫ +− )1t2t( 24t4 dt

= – ∫ +− )tt2t( 468 dt

= – 9

t9

+ 7

t2 7

– 5

t5

+ c

= – 9

xcos9

+ 2 7

xcos7

– 5

xcos5

+ c Ans.

Example : ∫− dx)x(cos)x(sin 3/73/1

Solution. ∫− dx)x(cos)x(sin 3/73/1

= ∫3/1)x(tan

xcos

12 dx

put tanx = t ⇒ sec2x dx = dt

= dtt 3/1

∫

= 4

3 t4/3 + c

= 4

3 (tanx)4/3 + c Ans.

Example : ∫ dxxcosxsin 42

Solution.8

1 ∫ + dx)x2cos1(x2sin2

= 8

1 ∫ dxx2sin2

+ 8

1 ∫ dxx2cosx2sin2

= 16

1 ∫ +−

16

1dx)x4cos1(

3

x2sin3

= 16

1 –

64

x4sin +

48

x2sin3

+ c

11 .11 .11 .11 . Integration of typeIntegration of typeIntegration of typeIntegration of type

∫ 1111xxxxKKKKxxxx 1111xxxx 22224444 2222++

± dx where K is any constant.

Divide Nr & Dr by x² & put x ∓ x

1 = t.

Example : ∫ 42

2

xx1

x1

++

−dx

Solution. ∫ 1x

1x

dxx

11

2

2

2

++

−

x + x

1 = t

⇒ – ∫ 1t

dt2 −

– 2

1 �n

1t

1t

+

− + C

– 2

1 �n

1x

1x

1x

1x

++

−+

+ C

Example : Evaluate : ∫ + 1x

14 dx

Solution. We have,

Ι = ∫ + 1x

14 dx

⇒ Ι = ∫+

2

2

2

x

1x

x

1

dx

⇒ Ι = 2

1 ∫

+2

2

2

x

1x

x

2

dx

⇒ Ι = 2

1 ∫

+

−

−

+

+

2

2

2

2

2

2

x

1x

x

11

x

1x

x

11

dx

⇒ Ι = 2

1 ∫

+

+

2

2

2

x

1x

x

11

dx – 2

1 ∫

+

−

2

2

2

x

1x

x

11

dx

⇒ Ι = 2

1 ∫

+

−

+

2x

1x

x

11

2

2

dx – 2

1 ∫

−

+

−

2x

1x

x

11

2

2

dx

Putting x – x

1 = u in 1st integral and x +

x

1 = ν in 2nd integral, we get

Ι = 2

1

( )∫+

22 2u

du

– 2

1 ( )∫

−ν

ν22 2

d

= 22

1 tan–1

2

u –

2

1

22

1 log

2

2

+ν

−ν + C

= 22

1 tan–1

−

2

x/1x –

24

1 log

2x/1x

2x/1x

++

−+ + C

= 22

1 tan–1

−

x2

1x2

– 24

1 log

12xx

1x2x2

2

++

+− + C

Self Practice Problem :

1. ∫ 1x7x

1x24

2

+−

− dx Ans.

6

1 �n

3x

1x

3x

1x

++

−+

+ C

2. ∫ xtan dx Ans.2

1 tan–1

2

y +

22

1 �n

2y

2y

+

− + C where y = tan x –

xtan

1

1 2 .1 2 .1 2 .1 2 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type

∫ ++ qqqqpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx OR ( )∫ +++ qqqqpxpxpxpxccccbxbxbxbxaxaxaxax dxdxdxdx2222 ; put px + q = t2.

Example: Evaluate : ∫ +− 1x)3x(

1 dx

Solution. Let Ι = ∫ +− 1x)3x(

1 dx

Here, P and Q both are linear, so we put Q = t2 i.e. x + 1 = t2 and dx = 2t dt

∴ Ι = 22

t

t2

)31t(

1∫ −−

dt

⇒ Ι = 2 ∫ − 22 2t

dt = 2 .

)2(2

1 log

2t

2t

+

− + C

⇒ Ι = 2

1 log

21x

21x

++

−+ + C.

Example : Evaluate : ∫ +++

+

1x)3x3x(

2x2 dx

Solution. Let Ι = ∫ +++

+

1x)3x3x(

2x2 dx

Putting x + 1 = t2, and dx = 2t dt, we get Ι = ∫+−+−

+

2222

2

t}3)1t(3)1t{(

dtt2)1t(

⇒ Ι = 2 ∫ ++

+

1tt

)1t(24

2

dt = 2 ∫++

+

1t

1t

t

11

2

2

2

dt

⇒ Ι = 2 ( )∫ ′+

22 3u

du

where t – t

1 = u.

⇒ Ι = 3

2 tan–1

3

u + C =

3

2 tan–1

−

3

t

1t

+ C

⇒ Ι = 3

2 tan–1

−

3t

1t2

+ C = 3

2 tan–1

+ )1x(3

x + C

13 .13 .13 .13 . Integration of typeIntegration of typeIntegration of typeIntegration of type

∫+++ rrrrqxqxqxqxpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx 2222 , put ax + b = tttt1111 ; ∫

++ qqqqpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx 22222222111121

Solution = ∫+

−

−

t

11

t

1

t

22

dt= ∫

+−

−

1t

1

t

1t

2

dt= ∫

+−

−

1tt2

+

−

−

4

3

2

1t

dtdt= ∫

1t

Example : ∫(x + ) x + x + 1dx , put x = tttt

= – �n

+

−+−

4

3

2

1t

2

1t

2

+ C

Example : ∫−+ 22 x1)x1(

dxSolution. Put x =

t

1 ⇒ Ι = ∫

−+ 1t1t 22 )(

dtput t2 – 1 = y2

⇒ Ι = – ∫ + y)2y( 2

dyy= –

2

1 tan–1

2

y + C

= – 2

1 tan–1

−

x2

x1 2

+ C

Self Practice Problems :

1. ∫ ++ 1x)2x(

dxAns. 2 tan–1 ( )1x + + C

2. ∫ +++ 1x)6x5x(

dx2 Ans. 2 tan–1 ( )1x + 2 tan–1

+

2

1x + C

3. ∫−++ 2xx1)1x(

dxAns. sin–1

+−

2

5

1x

1

2

3

+ C

4. ∫−+ 22 x1)1x2(

dxAns. –

3

1 tan–1

−2

2

x3

x1 + C

5. ∫−+++ 4x2x)2x2x(

dx

22 Ans. – 62

1�n

++−+

+−−+

)1x(64x2x

)1x(64x2x

2

2

+ C

1 4 .1 4 .1 4 .1 4 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type

∫ −

− xxxxββββ ααααxxxx dx or ( )( )∫ −− xxxxββββααααxxxx ; put x = αααα cos2 θθθθ + ββββ sin2 θθθθ

∫ −

−ββββxxxx ααααxxxx dx or ( )( )∫ −− ββββxxxxααααxxxx ; put x = αααα sec2 θ −θ −θ −θ − ββββ tan2 θθθθ

( )( )∫ −− ββββxxxxααααxxxx dxdxdxdx; put x −−−− αααα = t2 or x −−−− ββββ = t2.

1 5 .1 5 .1 5 .1 5 . Reduct ion formula of Reduct ion formula of Reduct ion formula of Reduct ion formula of ∫ dxxntan , , , , ∫ dxxncot , , , , ∫ dxxnsec ,,,,

∫ dxxecncos

1. Ιn = ∫ dxxtann

= ∫ − dxxtanxtan 2n2 = ∫ − )1x(sec2

tann – 2x dx

⇒ Ιn = ∫

−2n2 tanxsec + dx – Ιn – 2

⇒ Ιn =

1n

xtan 1n

−

−

– Ιn – 2

2. Ιn = ∫ dxxcotn

= ∫− dxxcot.cot 2n2

= ∫−− dxxcot)1xec(cos 2n2

⇒ Ιn = ∫

− dxxcotxeccos 2n2 – Ι

n – 2

⇒ Ιn = –

1n

xcot 1n

−

−

– Ιn – 2

3. Ιn = ∫ dxxsecn

= ∫− dxxsecxsec 2n2

⇒ Ιn = tanx secn – 2x – ∫ − )2n)(x(tan secn – 3 x. secx tanx dx.

⇒ Ιn = tanx secn – 2 x dx – (n – 2) (sec2 x – 1) secn – 2x dx

⇒ (n – 1) Ιn = tanx secn – 2x + (n – 2) Ι

n – 2

Ιn =

1n

xsecxtan 2n

−

−

+ 1n

2n

−

− Ι

n – 2

4. Ιn = ∫ dxeccos n

= ∫ xeccos 2 cosecn – 2 x dx

⇒ Ιn = – cotx cosecn – 2x + ∫ − )2n)(x(cot (– cosecn – 3x cosec x cot x) dx

⇒ – cotx cosecn – 2x – (n – 2) ∫− dxxeccosxcot 2n2

⇒ Ιn = – cotx cosecn – 2x – (n – 2) ∫ − )1xec(cos 2

cosecn – 2 x dx

⇒ (n – 1) Ιn = – cotx cosecn – 2 x + (n – 2) 2

n – 2

Ιn =

1n

xeccosxcot 2n

−

−

+ 1n

2n

−

− Ι

n – 2

Example : Obtain reducation formula for Ιn = ∫ sinnx dx. Hence evaluate ∫ sin4x dx

Solution. Ιn = ∫ (sin x) (sin x)n –1 dx

ΙΙ Ι

= – cos x (sin x)n–1 + (n – 1) ∫ (sin x)n–2 cos2x dx

= – cos x (sin x)n–1 + (n – 1) ∫ (sin x)n–2 (1 – sin2x) dx

Ιn = – cos x (sin x)n–1 + (n – 1) Ι

n–2 – (n – 1) Ι

n

⇒ Ιn = –

n

)x(sinxcos 1n−

+ n

)1n( − Ι

n–2(n ≥ 2)

Hence Ι4 = –

4

)x(sinxcos 3

+ 4

3

+− x

2

1

2

)x(sinxcos + C


1. ∫ 4x

3x

−

− dx Ans. )4x)(3x( −− + �n ( )4x3x −+− + C

2. ∫ 2/3)]x2)(1x[(

dx

−−Ans. 8

−

−−

−

−

1x

x2

x2

1x + C

3. ∫ 7/168 ])1x()2x[(

dx

−+Ans. 7

7/1

2x

1x

+

− + C

4. Deduce the reduction formula for Ιn = ∫ n4 )x1(

dx

+ and Hence evaluate Ι

2 = ∫ 24 )x1(

dx

+

Ans. Ιn = 1n4 )x1)(1n(4

x−+−

+ )1n(4

5n4

−

− Ι

n–1

Ι2 =

)x1(4

x4+

+ 4

3

++

−+−

−

−

2x

1x

2x

1x

n24

1

2

x

1x

tan22

1 1� + C

5. If Ιm,n

= ∫ (sin x)m (cos x)n dx then prove that

Ιm,n

= nm

)x(cos)x(sin 1n1m

+

−+

+ nm

1n

+

− . Ι

m,n–2

Definite IntegralsPART A :

A Let f(x) be a continuous function defined on [a, b],

∫ dx)x(f = F(x) + c. Then ∫b

a

dx)x(f = F(b) – F(a) is called definite integral. This formula is known as Newton-

Leibnitz formula.

Note :

1. The indefinite integral ∫ dx)x(f is a function of x, where as definite integral ∫b

a

dx)x(f is a number..

2. Given ∫ dx)x(f we can find ∫b

a

dx)x(f , but given ∫b

a

dx)x(f we cannot find ∫ dx)x(f

Illustration. 1 Evaluate ∫ ++

2

1)2x)(1x(

dx

Sol. ∵)2x)(1x(

1

++ =

1x

1

+ –

2x

1

+(by partial fractions)

∫ ++

2

1)2x)(1x(

dx = [ ]2

1ee )2x(log)1x(log +−+

= loge

3 – loge

4 – loge

2 + loge

3 =

8

9

elog


Evaluate the following

1. ∫ ++

2

1

2

2

3x4x

x5 dx Ans. 5 –

2

5

− 2

3

e4

5

e loglog9

2. ( )∫

π

++

2

0

32 2xxsec2 dx Ans.1024

4π +

2

π + 2

3. ∫

π

+

3

0xsec1

x dx Ans.

18

2π –

33

π + 2 log

e

3

2

PART B :Properties of definite integral

P – 1 ∫b

a

)x(f dx = ∫b

a

)t(f dt

i.e. definite integral is independent of variable of integration.

P – 2 ∫b

a

)x(f dx = – ∫a

b

)x(f dx

P – 3 ∫b

a

)x(f dx = ∫c

a

)x(f dx + ∫b

c

)x(f dx, where c may lie inside or outside the interval [a, b].

Illustration 2 If f(x) =

≥+

<+

3x:1x3

3x:3x2 , then find ∫

5

2

)x(f dx

Sol. ∫5

2

)x(f dx = ∫3

2

)x(f dx + ∫5

3

)x(f dx

= ∫ +

3

2

)3x( dx + ∫ +

5

3

2 )1x3( dx

= 2

49 − + 3 (3 – 2) + 53 – 33 + 5 – 3 =

2

211

Illustration 3 Evaluate ∫ −

8

2

|5x| dx

Sol. ∫ −

8

2

|5x| dx = ∫ +−

5

2

)5x( dx + ∫ +

8

5

)5x( dx = 9

Illustration 4 Show that ∫ +

2

0

)1x2( dx = ∫ +

5

0

)1x2( + ∫ +

2

5

)1x2(

Sol. L.H.S. = x2 + x ]20 = 4 + 2 = 6

R.H.S. = 25 + 5 – 0 + (4 + 2) – (25 + 5) = 6∴ L.H.S. R.H.S



1. ∫ −+

2

0

2 |3x2x| dx Ans. 4

2. ∫3

0

]x[ dx , where [x] is integral part of x. Ans. 3

3. [ ]∫9

0

t dt Ans. 13

PART C :

P – 4 ∫−

a

a

)x(f dx = ∫ −+

a

0

))x(f)x(f( dx

= 2 ∫a

0

)x(f dx if f(–x) = f(x) i.e. f(x) is even

= 0 if f(–x) = –f (x) i.e. f(x) is odd

Illustration 5 Evaluate ∫−

−

+

+1

1

x

xx

e1

ee dx

Sol. ∫−

−

+

+1

1

x

xx

e1

ee dx = ∫

+

++

+

+−

−−1

0

x

xx

x

xx

e1

ee

e1

ee dx

= ∫

+

++

+

+ −−1

0

x

xxx

x

xx

1e

)ee(e

e1

ee dx = ∫ −+

1

0

xx )ee( dx = e – 1 + 1

)1e( 1

−

−−

= e

1e2 −

Illustration 6 Evaluate ∫

π

π−

2

2

xcos dx

Sol. ∫

π

π−

2

2

xcos dx = 2 ∫

π

2

0

xcos dx = 2 (∵ cos x is even function)


+

−1

1

ex2

x2log dx

Sol. Let f(x) = loge

+

−

x2

x2

⇒ f(–x) = loge

−

+

x2

x2 = – log

e

+

−

x2

x2 = – f(x)

i.e. f(x) is odd function

∴ ∫−

+

−1

1

ex2

x2log dx = 0



1. ∫−

1

1

|x| dx Ans. 1

2. ∫

π

π−

2

2

7 xsin dx Ans. 0

3. ∫

π

π−

+

2

2

xe1

xcos dx Ans. 1

PART D :

P – 5 ∫b

a

)x(f dx = ∫ −+

b

a

)xba(f dx

Further ∫a

0

)x(f dx = ∫ −

a

0

)xa(f dx

Illustration 8 Prove that ∫

π

+

2

0)x(cosg)x(sing

)x(sing dx = ∫

π

+

2

0)x(cosg)x(sing

)x(cosg dx =

4

π

Sol. Let Ι = ∫

π

+

2

0)x(cosg)x(sing

)x(sing dx

⇒ Ι = ∫

π

−

π+

−

π

−

π2

0 x2

cosgx2

sing

x2

sing

= ∫

π

+

2

0)x(sing)x(cosg

)x(cosg dx

on adding, we obtain

2Ι = ∫

π

++

+

2

0)x(sing)x(cosg

)x(cosg

)x(cosg)x(sing

)x(sing dx = ∫

π

2

0

dx ⇒ Ι = 4

π

Note : 1. The above illustration can be remembered as a formula

2. Other similar formulae are

∫

π

+

2

0)x(cotg)x(tang

)x(tang dx = ∫

π

+

2

0)x(cotg)x(tang

)x(cotg dx =

4

π

∫

π

+

2

0)x(secg)ecx(cosg

)ecx(cosg dx = ∫

π

+

2

0)x(secg)ecx(cosg

)x(secg dx =

4

π

∫ −+

a

0)xa(g)x(g

)x(g dx =

2

a



1. ∫π

+0

xsin1

x dx Ans. π

2. ∫

π

+

2

0xcosxsin

x dx Ans.

22

π log

e ( )21+

3. ∫

π

+

2

0

44 xcosxsin

xcosxsinx dx Ans.

16

2π

4. ∫

π

π+

3

6

xtan1

dx

Ans.12

π

PART E :

P – 6 ∫a2

0

)x(f dx = ∫ −+

a

0

))xa2(f)x(f( dx

= 2 ∫a

0

)x(f dx if f (2a – x) = f(x)

= 0 if f (2a – x) = –f(x)

Illustration 9 Evaluate ∫π

0

33 xcosxsin dx

Sol. Let f(x) = sin3x cos3x ⇒ f(π – x) = – f(x)

∴ ∫π

0

33 xcosxsin dx = 0

Illustration 10 Evaluate ∫π

+0

2 xsin21

dx dx

Sol. Let f(x) = xsin21

12+

⇒ f(π – x) = f(x)

⇒ ∫π

+0

2 xsin21

dx = 2 ∫

π

+

2

0

2 xsin21

dx = 2 ∫

π

++

2

0

22

2

xtan2xtan1

dxxsec

= 2 ∫

π

+

2

0

2

2

xtan31

dxxsec =

3

2 ( )[ ]2

01 xtan3tan

π

−

∵ tan 2

π is undefined, we take limit

= 3

2

( ) ( )

− −−

π→

−0tan3tanxtan3tanLt 11

2x

= 3

2

2

π =

3

π

Note : We can evaluate the integral without using this property

Alternatively : ∫π

+0

2 xsin21

dx = ∫

π

+0

2

2

2xeccos

xeccos dx = ∫

π

+0

2

2

3xcot

dxxeccos

Observe that we are not converting in terms of tan x as it is not continuous in (0, π)

= – 3

1

π

−

0

1

3

xcottan = –

3

1

−

−

→

−

π→ +− 3

xcottanLt

3

xcottanLt 1

0x

1

x

= – 3

1

π−

π−

22 =

3

π

Note : If we convert in terms of tan x, then we have to break integral using property P – 3.

Illustration 11 Prove that ∫

π

2

0

e xsinlog dx = ∫

π

2

0

e xcoslog dx = ∫

π

2

0

e )x2(sinlog dx = – 2

π log

e

2 .

Sol. Let Ι = ∫

π

2

0

e xsinlog dx ..........(i)

⇒ Ι = ∫

π

−

π2

0

e x2

sinlog dx (by property P – 5)

Ι = ∫

π

2

0

e )x(coslog dx ..........(ii)

Adding (i) and (ii)

2 Ι = ∫

π

2

0

e )xcos.x(sinlog dx = ∫

π

2

0

e2

x2sinlog dx

2 Ι = ∫

π

2

0

e )x2(sinlog dx – ∫

π

2

0

2elog dx

2 Ι = Ι1 –

2

π log2

e..........(iii)

where Ι1 = ∫

π

2

0

e )x2(sinlog dx

put 2x = t ⇒ dx = 2

1 dt

L . L : x = 0 ⇒ t = 0

U . L : x = 2

π⇒ t = π

⇒ Ι1 = ∫

π

0

e )t(sinlog 2

1 dt

= 2

1 × 2 ∫

π

2

0

e )t(sinlog dt (by using property P – 6)

⇒ Ι1 = Ι ∴ (iii) gives Ι = –

2

π 2

elog



1. ∫∞

+

+

0

2

e

x1

x

1xlog

dx : Ans : π loge2

2. ∫−1

0

1

x

xsin dx : Ans :

2

π log

e2

3. ∫π

0

e xsinlogx dx Ans : –2

2π log

e2

PART F :

P – 7 If f(x) is a periodic function with period T, then

(i) ∫nT

0

)x(f dx = n ∫T

0

)x(f dx, n ∈ z

(ii) ∫+nTa

a

)x(f dx = n ∫T

0

)x(f dx, n ∈ z, a ∈ R

(iii) ∫nT

mT

)x(f dx = (n – m) ∫T

0

)x(f dx, m, n ∈ z

(iv) ∫+nTa

nT

)x(f dx = ∫a

0

)x(f dx, n ∈ z, a ∈ R

(v) ∫+

+

nTb

nTa

)x(f dx = ∫a

a

)x(f dx, n ∈ z, a, b ∈ R


2

1

}x{e dx

Sol. ∫−

2

1

}x{e dx = ∫+−

−

31

1

}x{e dx = 3 ∫1

0

}x{e dx = 3 ∫1

0

}x{e dx = 3(e – 1)

Illustration 13 Evaluate ∫+π vn

0

|xcos| dx , 2

π < v < π and n ∈ z

Sol. ∫+π vn

0

|xcos| dx = ∫v

0

|xcos| dx + ∫+π vn

v

|xcos| dx

= ∫

π

2

0

xcos – ∫π

v

xcos dx + n ∫π

0

|xcos| dx

= (1 – 0) – (sin v – 1) + 2n ∫

π

2

0

xcos dx

= 2 – sin v + 2n (1 – 0) = 2n + 2 – sin v

Self Practice Problem


1. ∫−

2

1

}x3{e dx Ans. 3 (e – 1)

2. ∫π

+

2000

0

xsine1

dx dx Ans. 1000 π

3. ∫

π

π+

4

5

44 xcosxsin

x2sin dx Ans.

4

π

PART G :

P – 8 If ψ(x) ≤ f(x) ≤ φ (x) for a ≤ x ≤ b, then

∫ψ

b

a

)x( dx ≤ ∫b

a

)x(f dx ≤ ∫φ

b

a

)x( dx

P – 9 If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m (b – a) ≤ ∫b

a

)x(f dx ≤ M (b – a)

Further if f(x) is monotonically decreasing in (a, b) then f(b) (b – a) < ∫b

a

)x(f dx < f(a) (b – a) and if f(x)

is monotonically increasing in (a, b) then f(a) (b – a) < ∫b

a

)x(f dx < f(b) (b – a)

P – 10 ∫b

a

dx)x(f ≤ ∫b

a

dx)x(f

P – 11 If f(x) ≥ 0 on [a, b] then ∫b

a

dx)x(f ≥ 0

Illustration 14 For x ∈ (0, 1) arrange f1(x) =

2x4

1

−, f

2(x) =

2x24

1

− and f

3(x) =

32 xx4

1

−− in ascending

order and hence prove that 6

π < ∫

−−

1

032 xx4

dx <

24

π

Sol. ∵ 0 < x3 < x2 ⇒ x2 < x2 + x3 < 2x2 ⇒ –2x2 < – x2 – x3 < –x2

⇒ 4 – 2x2 < 4 –x2 – x3 < 4 – x2

⇒ 2x24 − < 32 xx4 −− < 2x4 −

⇒ f1(x) < f

3(x) < f

2(x) for x ∈ (0, 1)

⇒ ∫1

0

1 )x(f dx < ∫1

0

3 )x(f dx < ∫1

0

2 )x(f dx

sin–1

1

02

x

< ∫

−−

1

032 xx4

dx <

2

1 sin–1

1

02

x

6

π < ∫

−−

1

032 xx4

dx <

24

π

Illustration 15 Estimate the value of ∫

π

2

0x

xsin dx

Sol. Let f(x) = x

xsin

f′(x) = 2x

xsinxcosx − = 2x

)xtanx)(x(cos − < 0

⇒ f(x) is monotonically decreasing function.f(0) is not defined, so we evaluate

+→0xLt f(x) = +→0x

Lt x

xsin = 1. Take f(0) = +→0x

Lt f(x) = 1

f

π

2 =

π

2

π

2 .

−

π0

2 < ∫

π

2

0x

xsin dx < 1 .

−

π0

2

1 < ∫

π

2

0x

xsin dx <

2

π

Note : Here by making the use of graph we can make more appropriate approximation as in next illustration.

Illustration 16 Estimate the value of ∫1

0

x dxe2

using (i) rectangle, (ii) triangle

Sol. (i) By using rectangle

Area OAED < ∫1

0

x dxe2

< Area OABC

1 < ∫1

0

x dxe2

< 1 . e

1 < ∫1

0

x dxe2

< e

(ii) By using triangle

Area OAED < ∫1

0

x dxe2

< Area OAED + Area of triangle DEB

1 < ∫1

0

x dxe2

< 1 + 2

1 . 1. (e – 1) 1 < ∫

1

0

x dxe2

< 2

1e +

Illustration 17 Estimate the value of ∫1

0

x dxe2

by using ∫1

0

xdxe

Sol. For x ∈ (0, 1), 2xe < ex

⇒ 1 × 1 < ∫1

0

x dxe2

< ∫1

0

xdxe

1 < ∫1

0

x dxe2

< e – 1

Exercise : Prove the following :

1. ∫−

1

0

2x xcose dx < ∫−

1

0

2x xcose2

dx

2. 0 < ∫

π

+2

0

1n xsin dx < ∫

π

2

0

2 xsin dx

3. 4

1

e−

< ∫−

1

0

xx2

e dx < 1

4. – 2

1 ≤ ∫ +

1

0

2

3

x2

xcosx dx <

2

1

5. 1 < ∫

π

2

0

xsin dx < 2

π

6. 0 < ∫ +

2

0

3x16

dxx <

6

1

PART - H

Leibnitz Theorem : If F(x) = ∫)x(h

)x(g

dt)t(f , then

dx

)x(dF = h′(x) f(h(x)) – g′(x) f(g(x))

Proof : Let P(t) = ∫ dt)t(f

⇒ F(x) = ∫)x(h

)x(g

dt)t(f = P(h(x)) – P(g(x))

⇒dx

)x(dF= P′(h(x)) h′(x) – P′(g(x)) g′(x)

= f(h(x)) h′(x) – f (g(x)) g′(x)

Illustration 18 If F(x) = dttsin

2x

x

∫ , then find F′(x)

Sol. F′(x) = 2x . 2xsin – 1 . xsin

Illustration 19 If F(x) = dtlog

tx3

x2

e

e

te

∫ , then find first and second derivative of F(x) with respect to logxe

at x = 2elog

Sol. ( )xelogd

)x(dF =

dx

)x(dF

)x(logd

dx =

−

x2x3 ee

x2x2

ee

x3x3

log

ee.2

log

e.e.3 x = e6x – e4x.

( )2xe

2

logd

)x(Fd = ( )x

elogd

d (e6x – e4x) =

dx

d (e6x – e4x) ×

dx

logd

1xe

= (6 e6x – 4 e4x) x

First derivative of F(x) at x = log2e (i.e. ex = 2) is 26 – 24 = 48

Second derivative of F(x) at x = log2e (i.e. ex = 2) is (6 . 26 – 4 . 24) . log

e2 = 5 . 26 . log2

e.

Illustration 20 Evaluate ∞→x

Lt

∫

∫

x

0

t2

2x

0

t

dte

dte

2

2

Sol.∞→x

Lt

∫

∫

x

0

t2

2x

0

t

dte

dte

2

2

∞

∞form

Applying L′ Hospital rule

=∞→x

Lt2

22

x2

x

0

xt

e.1

e.dte.2 ∫

=∞→x

Lt2

2

x

x

0

t

e

dte.2 ∫=

∞→xLt 2

2

x

x

e.x2

e.2 = 0

Modified Leibnitz Theorem :

If F(x) = ∫)x(h

)x(g

)t,x(f dt, then

FFFF ′(x) = ∫ ∂

∂)x(h

)x(gx

)t,x(f dt + f(x, h(x))h′(x) – f(x, g(x)) . g′(x)

Illustration 21 If f(x) = ∫ +

x

logxe

tx

dt

, then find f′(x)

Sol. f′(x) = ( )∫ +

−x

log

2xe

dttx

1 + 1 .

x2

1 –

x

1 ( )x

elogx

1

+ =

x

logxe

)tx(

1

+ + x2

1 – ( )x

elogxx

1

+

= x2

1 – x

elogx

1

+ +

x2

1 – ( )x

elogxx

1

+=

x

1 – ( )x

elogxx

1x

+

+ = ( )x

e

xe

logxx

1log

+

−

Alternatively : f(x) =

x

log

e

x

log xe

xe

)tx(logtx

dt

+=

+∫ (treating ‘t’ as constant)

f(x) = loge2x – log

e (x + log x

e)

f′(x) = x

1 – ( )x

elogx

1

+

+

x

11 = ( )x

e

xe

logxx

1log

+

−

Definite Integrals dependent on parameters :


1

0

xe

b

log

1x, ‘b’ being parameter

Sol. Let Ι(b) = ∫−

1

0

xe

b

log

1x dx

db

)b(dΙ = ∫

1

0

xe

xe

b

log

logx dx + 0 – 0

(using modified Leibnitz Theorem)

= ∫1

0

b dxx =

1

0

1b

1b

x

+

+

= 1b

1

+

Ι(b) = loge (b + 1) + c

b = 0 ⇒ Ι(0) = 0∴ c = 0 ∴ Ι(b) = log

e (b+1)


−1

02

1

x1x

)ax(tan dx , ‘a’ being parameter

Sol. Let Ι(a) = ∫−

−1

02

1

x1x

)ax(tan dx

da

)a(dΙ = ∫ +

1

0

22 )xa1(

x 2x1x

1

− dx = ∫

−+

1

0222 x1)xa1(

dx

Put x = sin t ⇒ dx = cos t dtL.L. : x = 0 ⇒ t = 0

U.L. : x = 1 ⇒ t = 2

π

da

)a(dΙ = ∫

π

+

2

0

22 tsina1

1

tcos

1 cos t dt = ∫

π

+

2

0

22 tsina1

dt

= ∫

π

++

2

0

22

2

ttan)a1(1

dttsec =

2a1

1

+ tan–1

2

0

2 ttana1

π

+

= 2a1

1

+ .

2

π

⇒ Ι(a) = 2

π log

e

++ 2a1a + c

But Ι(0) = 0 ⇒ c = 0

⇒ Ι(a) = 2

π log

e

++ 2a1a


1. If f(x) = ∫3x

0

tcos dt, find f′(x). Ans. 3x2 3xcos

2. If f(x) = eg(x) and g(x) = ∫ +

x

2

4t1

t dt then find the value of f′(2). Ans.

17

2

3. If x = ∫+

y

02t41

dt and 2

2

dx

yd = Ry then find R Ans. 4

4. If f(x) = ∫2x

x

2 dttsinx then find f′(x). Ans. x2 (2x sin x2 – sin x) + (cos x – cos x2) x

5. If φ(x) = cos x – ∫ φ−

x

0

)t()tx( dt, then find the value of φ′′(x) + φ(x). Ans. – cos x

6. Find the value of the function f(x) = 1 + x + ( )∫ +

x

1

te

2te log2)(log dt where f′(x) vanishes. Ans. 1 +

e

2

7. Evaluate 0x

Lt→

xsinx

dttcos

2x

0

2∫. Ans. 1

8. Evaluate ∫π

+

0

e )xcosb1(log dx, ‘b’ being parameter.. Ans. π loge

−+

2

b11 2

PART - ΙΙΙΙ

Definite Integral as a Limit of Sum.

Let f(x) be a continuous real valued function defined on the closed interval [a, b] which is divided into n parts

as shown in figure.

y = f(x)

a a+h a+2h ................ a+(n-1)h a+nh=bx

The point of division on x-axis are a, a + h, a + 2h ..........a + (n – 1)h, a + nh, where n

ab − = h.

Let Sn denotes the area of these n rectangles.

Then, Sn = hf(a) + hf(a + h) + hf(a + 2h) + ........+hf(a + (n – 1)h)

Clearly, Sn is area very close to the area of the region bounded by curve y = f(x), x–axis and the ordinates

x = a, x = b.

Hence ∫b

a

dx)x(f = ∞→n

Lt Sn

∫b

a

dx)x(f = ∞→n

Lt ∑−

=

+

1n

0r

)rha(fh = ∞→n

Lt ∑−

=

−1n

0rn

ab f

−+

n

r)ab(a

Note :

1. We can also write

Sn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and ∫

b

a

dx)x(f = ∞→n

Lt ∑=

−n

1rn

ab f

−+ r

n

aba

2. If a = 0, b = 1, ∫1

0

dx)x(f = ∞→nLt ∑

−

=

1n

0r n

rf

n

1

Steps to express the limit of sum as definte integral

Step 1. Replace n

r by x,

n

1 by dx and ∞→n

Lt ∑ by ∫

Step 2. Evaluate ∞→nLt

n

r by putting least and greatest values of r as lower and upper limits respectively..

For example ∞→nLt ∑

=

pn

1rn

rf

n

1 = ∫

p

0

dx)x(f (∵ 1r

n n

rLt

=∞→

= 0,

nprn n

rLt

=∞→

= p)

Illustration 25 : Evaluate

∞→nLt

++

++

++

+ n2

1.........

n3

1

n2

1

n1

1

Sol. ∞→nLt

++

++

++

+ n2

1.........

n3

1

n2

1

n1

1

= ∞→nLt ∑

=+

n

1rnr

1

= ∞→nLt ∑

=

n

1rn

1

1n

r

1

+

= ∫ +

1

01x

dx = [ ]10e )1x(log + = log

e2.

Illustration 26 : Evaluate ∞→nLt

++

+

++

+

++

+

+

n5

3.........

3n

3n

2n

2n

1n

1n222222

Sol. ∞→nLt ∑

= +

+n2

1r22 rn

rn = ∞→n

Lt ∑=

n2

1rn

1 2

n

r1

n

r1

+

+

∵ ∞→nLt

n

r = 0, when r = 1, lower limit = 0

and ∞→nLt

n

r = ∞→n

Lt

n

n2 = 2, when r = 2n, upper limit = 2

∫ +

+2

0

2x1

x1 dx = ∫ +

2

0

2x1

1 dx +

2

1 ∫ +

2

0

2x1

x2 dx

= tan–1x]20 +

2

0

2e )x1(log

2

1

+

= tan–1 2 + 2

1 log

e5

Illustration 27 : Evaluate

∞→nLt

n

1

nn

!n

Sol. Let y = ∞→nLt

n

1

nn

!n

loge y = ∞→n

Lt n

1 log

e

nn

!n

= ∞→nLt

n

1log

e

nn

n........3.2.1

= ∞→nLt

n

1

++

+

+

n

nlog.....

n

3log

n

2log

n

1log eeee

= ∞→nLt

n

1 ∑=

n

1r

en

rlog

=

1

0

e

1

0

e xxlogxdxxlog

−=∫

= (0 – 1) – +→0x

Lt x logex + 0

= – 1 – 0 = –1

⇒ y = e

1


Evaluate the following limits

1. ∞→nLt

+++

++

++

22222 nn

1....

n2n

1

nn

1

n

1Ans. 2 ( )12 −

2. ∞→nLt

++

++

++

+ n5

1.......

n3

1

n2

1

n1

1Ans. log

e5

3. ∞→nLt 2n

1

π++

π+

π+

π

n4

nsinn........

n4

3sin3

n4

2sin2

n4sin 3333

Ans. 29

2

π (52 – 15π)

4. ∞→nLt ∑

−

=

1n

0r22 rn

1

−Ans.

2

π

5. ∞→nLt

n

3

−+++

++

++

++

)1n(3n

n......

9n

n

6n

n

3n

n1 Ans. 2

PART – J

Reduction Formulae in Definite Integrals

1. If Ιn = ∫

π

2

0

n xsin dx , then show that Ιn =

−

n

1n Ι

n–2

Proof : Ιn = ∫

π

2

0

n xsin dx

Ιn = [ ]201n xcosxsin

π

−− + ∫

π

−−

2

0

22n dxxcos.xsin)1n(

= (n – 1) ∫

π

− −

2

0

22n dx)xsin1(.xsin

= (n – 1) ∫

π

− −−

2

0

2n )1n(dxxsin ∫

π

2

0

n dxxsin

Ιn + (n – 1) Ι

n = (n – 1) Ι

n–2

Ιn =

−

n

1n Ι

n–2

Note : 1. ∫

π

2

0

n dxxsin = ∫

π

2

0

n dxxcos

2. Ιn =

−

n

1n

−

−

2n

3n

−

−

4n

5n ..... Ι

0 or Ι

1

according as n is even or odd. Ι0 =

2

π, Ι

1 = 1

Hence Ιn =

−

−

−

−

−

π

−

−

−

−

−

oddisnif1.3

2........

4n

5n

2n

3n

n

1n

evenisnif2

.2

1........

4n

5n

2n

3n

n

1n

2. If Ιn = ∫

π

4

0

n dxxtan , then show that Ιn + Ι

n–2 =

1n

1

−

Sol. Ιn = ∫

π

−4

0

2n)x(tan . tan2x dx

= ∫

π

−4

0

2n)x(tan (sec2x – 1) dx

= ∫

π

−4

0

2n)x(tan sec2x dx – ∫

π

−4

0

2n)x(tan dx

= 4

0

1n

1n

)x(tan

π−

− – Ι

n–2

Ιn =

1n

1

− – Ι

n–2

∴ Ιn + Ι

n–2 =

1n

1

−

3. If Ιm,n

= ∫

π

2

0

nm dxxcos.xsin , then show that Ιm,n

= nm

1m

+

− Ι

m–2 , n

Sol. Ιm,n

= ∫

π

−2

0

n1m )xcosx(sinxsin dx

= 2

0

1n1m

1n

xcos.xsin

π+−

+− + ∫

π

+

+

2

0

1n

1n

xcos (m – 1) sinm–2 x cos x dx

=

+

−

1n

1m ∫

π

−2

0

2n2m xcos.xcos.xsin dx

=

+

−

1n

1m ( )∫

π

− −

2

0

nmn2m xcos.xsinxcos.xsin dx

=

+

−

1n

1m Ι

m–2,n –

+

−

1n

1m Ι

m,n

⇒

+

−+

1n

1m1 Ι

m,n =

+

−

1n

1m Ι

m–2,n

Ιm,n

=

+

−

nm

1m Ι

m–2,n

Note : 1. Ιm,n

=

+

−

nm

1m

−+

−

2nm

3m

−+

−

4nm

5m ........ Ι

0,n or Ι

1,n according as m is even or odd.

Ι0,n

= ∫

π

2

0

n dxxcos and Ι1,n

= ∫

π

2

0

n dxxcos.xsin = 1n

1

+

2. Walli’s Formula

Ιm,n

=

−+−++

−−−−−−

π

−+−++

−−−−−−

otherwise)........4nm()2nm()nm(

).......5n()3n()1n.........()5m()3m()1m(

evenaren,mbothwhen2)........4nm()2nm()nm(

).......5n()3n()1n.........()5m()3n()1n(

Illustration 28 : Evaluate ∫

π

π−

+2

2

22 dx)xcosx(sinxcosxsin

Sol. Given integral = ∫

π

π−

2

2

23 dxxcosxsin + ∫

π

π−

2

2

32 dxxcosxsin

= 0 + 2 ∫

π

2

0

32 dxxcosxsin (∵ sin3x cos2x is odd and sin2x cos3x is even)

= 2. 1.3.5

2.1 =

15

4

Illustration 29 : Evaluate ∫π

0

65 dxxcosxsinx

Sol. Let Ι = ∫π

0

65 dxxcosxsinx

Ι = ∫π

−π−π−π

0

65 dx)x(cos)x(sin)x(

= π ∫π

0

65 dxxcos.sin – ∫π

0

65 dxxcos.xsinx

⇒ 2Ι = π . 2 ∫

π

2

0

65 dxxcos.xsin

Ι = π 1.3.5.7.9.11

1.3.5.2.4

Ι = 693

8π

Illustration 30 : Evaluate ∫ −

1

0

53 dx)x1(x

Sol. Put x = sin2θ ⇒ dx = 2 sin θ cos θ dθL . L : x = 0 ⇒ θ = 0

U.L. : x = 1 ⇒ θ = 2

π

∴ ∫ −

1

0

53 dx)x1(x = ∫

π

θθ

2

0

526 )(cossin 2 . sin θ . cos θ dθ

= 2 . ∫

π

θθθ

2

0

117 dcossin

= 2 . 2.4.6.8.10.12.14.16.18

2.4.6.8.10.2.4.6=

504

1

Self Practice Problems:


1. ∫

π

2

0

5 dxxsin Ans.15

18

2. ∫

π

2

0

45 dxxcosxsin Ans.315

8

3. ∫ −

1

0

16 dxxsinx Ans.14

π –

245

16

4. ( )∫ −

a

0

2

722 dxxax Ans.

9

a9

5. ∫ −

2

0

2/3 dxx2x Ans.2

π

KEY CONCEPTS1. DEFINITION :

If f & g are functions of x such that g′(x) = f(x) then the function g is called a PRIMITIVE OR

ANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as

∫ f(x) dx = g(x) + c ⇔ d

dx{g(x) + c} = f(x), where c is called the constant of integration.

2. STANDARD RESULTS :

(i) ∫ (ax + b)n dx = ( )

( )ax b

a n

n++

+1

1 + c n ≠ −1 (ii) ∫ dx

ax b+ =

1

a ln (ax + b) + c

(iii) ∫ eax+b dx = 1

a eax+b + c (iv) ∫ apx+q dx =

1

p

a

na

px q+

� (a > 0) + c

(v) ∫ sin (ax + b) dx = − 1

a cos (ax + b) + c (vi) ∫ cos (ax + b) dx =

1

a sin (ax + b) + c

(vii) ∫ tan(ax + b) dx = 1

a

ln sec (ax + b) + c (viii) ∫ cot(ax + b) dx = 1

a ln sin(ax + b)+ c

(ix) ∫ sec² (ax + b) dx = 1

a tan(ax + b) + c (x) ∫ cosec²(ax + b) dx = −

1

acot(ax + b)+ c

(xi) ∫ sec (ax + b) . tan (ax + b) dx = a

1 sec (ax + b) + c

(xii) ∫ cosec (ax + b) . cot (ax + b) dx = a

1− cosec (ax + b) + c

(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π4 2

+

x+ c

(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tan 2

x + c OR − ln (cosecx + cotx)

(xv) ∫ sinh x dx = cosh x + c (xvi) ∫ cosh x dx = sinh x + c (xvii) ∫ sech²x dx = tanh x + c

(xviii) ∫ cosech²x dx = − coth x + c (xix) ∫ sech x . tanh x dx = − sech x + c

(xx) ∫ cosech x . coth x dx = − cosech x + c (xxi) ∫22 xa

xd

− = sin−1

a

x + c

(xxii) ∫ 22 xa

xd

+ =

a

1 tan−1

a

x + c (xxiii) ∫ 22 axx

xd

− =

a

1 sec−1

a

x + c

(xxiv) ∫ d x

x a2 2+ = ln

++ 22 axx OR sinh−1

a

x + c

(xxv) ∫22 ax

xd

− = ln

−+ 22 axx OR cosh−1

a

x + c

(xxvi) ∫ 22 xa

xd

− =

a2

1 ln

xa

xa

−+

+ c (xxvii) ∫ 22ax

xd

− =

a2

1 ln

ax

ax

+−

+ c

(xxviii) ∫ 22xa − dx =

2

x 22

xa − + 2

a2

sin−1

a

x + c

(xxix) ∫ 22 ax + dx = 2

x 22 ax + +

2

a2

sinh−1

a

x + c

(xxx) ∫ 22 ax − dx = 2

x 22 ax − −

2

a2

cosh−1

a

x + c

(xxxi) ∫ eax. sin bx dx = 22

ax

ba

e

+ (a sin bx − b cos bx) + c

(xxxii) ∫ eax . cos bx dx = 22

ax

ba

e

+ (a cos bx + b sin bx) + c

3. TECHNIQUES OF INTEGRATION :(i) Substitution or change of independent variable .

Integral I = ∫ f(x) dx is changed to ∫ f(φ (t)) f ′ (t) dt , by a suitable substitution

x = φ (t) provided the later integral is easier to integrate .

(ii) Integration by part : ∫ u.v dx = u ∫ v dx − ∫

∫ xdv.

xd

ud dx where u & v are differentiable

function . Note : While using integration by parts, choose u & v such that

(a) ∫ v dx is simple & (b) ∫

∫ xdv.

xd

ud dx is simple to integrate.

This is generally obtained, by keeping the order of u & v as per the order of the letters in ILATE,where ; I − Inverse function, L − Logarithmic function ,A − Algebraic function, T − Trigonometric function & E − Exponential function

(iii) Partial fraction , spiliting a bigger fraction into smaller fraction by known methods .

4. INTEGRALS OF THE TYPE :

(i) ∫ [ f(x)]n f ′(x) dx OR ∫ [ ]n)x(f

)x(f ′ dx put f(x) = t & proceed .

(ii)dx

ax bx c2 + +∫ , dx

ax bx c2 + +∫ , ax bx c2 + +∫ dx

Express ax2 + bx + c in the form of perfect square & then apply the standard results .

(iii)px q

ax bx c

++ +∫ 2 dx ,

px q

ax bx c

+

+ +∫ 2

dx .

Express px + q = A (differential co-efficient of denominator) + B .

(iv) ∫ ex [f(x) + f ′(x)] dx = ex . f(x) + c (v) ∫ [f(x) + xf ′(x)] dx = x f(x) + c

(vi) ∫)1x(x

xdn +

n ∈ N Take xn common & put 1 + x−n = t .

(vii) ∫( ) n

)1n(n2 1xx

xd−

+ n ∈ N , take xn common & put 1+x−n = tn

(viii)( )

dx

x xn nn

11

+∫ /

take xn common as x and put 1 + x −n = t .

(ix) ∫xsinba

xd2+

OR ∫xcosba

xd2+

OR ∫xcoscxcosxsinbxsina

xd22 ++

Multiply ..rN & ..rD by sec² x & put tan x = t .

(x) ∫xsinba

xd

+ OR ∫

xcosba

xd

+ OR ∫

xcoscxsinba

xd

++

Hint :Convert sines & cosines into their respective tangents of half the angles , put tan 2

x = t

(xi) ∫nxsin.mxcos.

cxsin.bxcos.a

++++

� dx . Express Nr ≡ A(Dr) + B

xd

d (Dr) + c & proceed .

(xii) ∫1xKx

1x24

2

+++

dx OR ∫1xKx

1x24

2

++−

dx where K is any constant .

Hint : Divide Nr & Dr by x² & proceed .

(xiii)dx

ax b px q( )+ +∫ & ( )dx

ax bx c px q2 + + +∫ ; put px + q = t2 .

(xiv) dx

ax b px qx r( )+ + +∫ 2

, put ax + b = 1

t ;

( )dx

ax bx c px qx r2 2+ + + +∫ , put x =

1

t

(xv)x

x

−−∫

αβ

dx or ( ) ( )x x− −∫ α β ; put x = α cos2 θ + β sin2 θ

x

x

−−∫

αβ

dx or ( ) ( )x x− −∫ α β ; put x = α sec2 θ − β tan2 θ

( ) ( )dx

x x− −∫

α β ; put x − α = t2 or x − β = t2 .

DEFINITE INTEGRAL

1. ∫b

a

f(x) dx = F(b) − F(a) where ∫ f(x) dx = F(x) + c

VERY IMPORTANT NOTE : If ∫b

a

f(x) dx = 0 ⇒ then the equation f(x) = 0 has atleast one

root lying in (a , b) provided f is a continuous function in (a , b) .

2. PROPERTIES OF DEFINITE INTEGRAL :

P−−−−1 ∫b

a

f(x) dx = ∫b

a

f(t) dt provided f is same P −−−− 2 ∫b

a

f(x) dx = − ∫a

b

f(x) dx

P−−−−3 ∫b

a

f(x) dx = ∫c

a

f(x) dx + ∫b

c

f(x) dx , where c may lie inside or outside the interval [a, b] . This property

to be used when f is piecewise continuous in (a, b) .

P−−−−4 ∫−

a

af(x) dx = 0 if f(x) is an odd function i.e. f(x) = − f(−x) .

= 2 ∫a

0

f(x) dx if f(x) is an even function i.e. f(x) = f(−x) .

P−−−−5 ∫b

a

f(x) dx = ∫b

a

f(a + b − x) dx , In particular ∫a

0

f(x) dx = ∫a

0

f(a − x)dx

P−−−−6 ∫a2

0

f(x) dx = ∫a

0

f(x) dx + ∫a

0

f(2a − x) dx = 2 ∫a

0

f(x) dx if f(2a − x) = f(x)

= 0 if f(2a − x) = − f(x)

P−−−−7 ∫an

0

f(x) dx = n ∫a

0

f(x) dx ; where‘a’is the period of the function i.e. f(a + x) = f(x)

P−−−−8a nT

b nT

+

+

∫ f(x) dx = a

b

∫ f(x) dx where f(x) is periodic with period T & n ∈ I .

P−−−−9ma

na

∫ f(x) dx = (n − m) 0

a

∫ f(x) dx if f(x) is periodic with period 'a' .

P−−−−10 If f(x) ≤ φ(x) for a ≤ x ≤ b then ∫b

a

f(x) dx ≤ ∫b

a

φ (x) dx

P−−−−11 xd)x(fb

a

∫ ≤ ∫b

a

f(x)dx .

P−−−−12 If f(x) ≥ 0 on the interval [a, b] , then a

b

∫ f(x) dx ≥ 0.

3. WALLI’S FORMULA :

∫π 2/

0

sinnx . cosmx dx = [ ][ ]

2or1)....4nm()2nm()nm(

2or1)....3m()1m(2or1)....5n()3n()1n(

−+−++−−−−−

K

Where K = 2

π if both m and n are even (m, n ∈ N) ;

= 1 otherwise4. DERIVATIVE OF ANTIDERIVATIVE FUNCTION :

If h(x) & g(x) are differentiable functions of x then ,

xd

d ∫)x(h

)x(g

f(t) dt = f [h (x)] . h′(x) − f [g (x)] . g′(x)

5. DEFINITE INTEGRAL AS LIMIT OF A SUM :

∫b

a

f(x) dx = ∞→nLimit h [f (a) + f (a + h) + f (a + 2h) + ..... + f ( )a n h+ −1 ]

= 0hLimit

→ h ∑=

−

r

n

0

1

f (a + rh) where b − a = nh

If a = 0 & b = 1 then , ∞→nLimit h ∑

=

−

r

n

0

1

f (rh) = ∫1

0

f(x) dx ; where nh = 1 OR

∞→nLimit

n

1

1n

1r

−

=∑ f

n

r = ∫

1

0

f(x) dx .

6. ESTIMATION OF DEFINITE INTEGRAL :

(i) For a monotonic decreasing function in (a , b) ; f(b).(b − a) < ∫b

a

f(x) dx < f(a).(b − a) &

(ii) For a monotonic increasing function in (a , b) ; f(a).(b − a) < ∫b

a

f(x) dx < f(b).(b − a)

7. SOME IMPORTANT EXPANSIONS :

(i) 1 − 2ln.....5

1

4

1

3

1

2

1=∞++−+ (ii)

6.....

4

1

3

1

2

1

1

1 2

2222

π=∞++++

(iii)12

.....4

1

3

1

2

1

1

1 2

2222

π=∞+−+− (iv)

8.....

7

1

5

1

3

1

1

1 2

2222

π=∞++++

(v)24

.....8

1

6

1

4

1

2

1 2

2222

π=∞++++

EXERCISE–1

Q.1 ∫ θθ+θ

θd

sincos

2tan

66Q.2

( )5 4

1

4 5

52

x x

x x

+

+ +∫ dx Q.3 ∫ +

dxxtan1

xcos2

Q.4

( )dx

x42

1−∫ Q.5 Integrate ∫

dx

x x x2 2 1+ − by the substitution z = x + x x

22 1+ −

Q.6 x

e

e

xnx dx

x x +

∫ � Q.7 ∫ cos 2θ . lnθ−θθ+θ

sincos

sincosdθ Q.8 ∫ x2sinxsin

xd2 +

Q.9 ∫ xcosbxsina

xcosbxsina2424

2222

++

dx Q.10 ( )∫++

2)x1(xx

dxQ.11 ∫ 2xx 2++ dx

Q.12 ∫ sin( )

sin( )

x a

x a

−+

dx Q.13 ∫ (sin x)−11/3 (cos x)−1/3dx Q.14 ∫ )1x(sec)xsin1(

dxxcot

+−

Q.15 ∫ 1

1

−+

x

x dx Q.16 ∫ sin−1

xa

x

+ dx

Q.17 ∫( )[ ]

−++4

22

x

xln21xnl1x dx Q.18 ∫

+

2)x(ln

1)x(lnnl dx Q.19 ∫ ( )2xex1x

1x

+

+ dx

Q.20 Integrate 1

2 f ′ (x) w.r.t. x4 , where f (x) = tan −1x + ln 1+ x − ln 1− x

Q.21 ∫ ++

)1x(x

dx)1x(3 Q.22

dx

x xsin cos2 2

3∫ Q.23 ∫ ++

+dx

)1xe(

xx2x

2

Q.24 ∫ xsec21

xsec.

xcotxeccos

xcotxeccos

++−

dx Q.25 ∫ x2sin97

xsinxcos

−−

dx Q.26 ∫xeccosxsec

xd

+ dx

Q.27 ∫ xsecxsin

xd

+ Q.28 ∫ tan x.tan 2x.tan 3x dx Q.29 ( )

dx

x xsin sin 2 +∫

α

Q.30 dx)xcosxsinx)(xsinxcosx(

x2

∫ +−Q.31 ∫ xcosxsin23

xcos2xsin43

++++

dx

Q.32 ∫( )�n x x

x

cos cos

sin

+ 2

2dx

Q.33 sin

sin cos

x

x x+∫ dx Q.34 ∫ xtanxsin

xd

+Q.35 ∫ −

+dx

)1x(

1x332

2

Q.36 ∫+

dxxsin

)xcosxsinx(e2

3xcos

Q.37 ∫( )

( )

ax b dx

x c x ax b

2

2 2 2 2

−

− +Q.38 ∫

( )2

2x

x1)x1(

x2e

−−

−dx

Q.39 ∫ 2/3)x10x7(

x

2−− dx Q.40 ∫ ( ) 2/32 1x

xlnx

− dx Q.41 ∫

1

13

−+

x

x

dx

x

Q.42 2 3

2 3

1

1

−+

+−∫

x

x

x

x dx Q.43

cot tan

sin

x x

x

−+∫

1 3 2 dx Q.44 ∫ +

−+−+−dx

)1x(x

7x4x2x8x7x4222

2345

Q.45 ∫ ( ) 1x

xd

3x3x

2x2 +++

+Q.46 ∫

2 2

2

− −x x

x dx Q.47 ∫ )x()x()x(

xd

β−α−α−

Q.48 ∫++−+ 1x4x6x4x

dxx

234Q.49 ∫ xsin

x2cos dx Q.50 ∫ +α−

+42

2

xcosx21

dx)x1(α∈(0, π)

EXERCISE–2

Q.1 ∫π

0

x dx

x x9 2 2cos sin+ Q.2 1 2

1 20

2 −+z sin

sin

x

xdx

π

Q.3 Evaluate In = ∫

e

1

(lnn x) dx hence find I3.

Q.4 ∫π 2/

0

sin2x · arc tan(sinx) dx Q.5 ∫π 2/

0

cos4 3x . sin2 6x dx Q.6 ∫π 4/

0

x dx

x x xcos (cos sin )+

Q.7 Let h (x) = (fog) (x) + K where K is any constant. If ( ))x(hdx

d = –

)x(coscos

xsin2

then compute the

value of j (0) where j (x) = ∫)x(f

)x(g

dt)t(g

)t(f, where f and g are trigonometric functions.

Q.8 Find the value of the definite integral ∫π

+0

dxxcos2xsin2 .

Q.9 Evaluate the integral : x x x x+ − + − −

∫ 2 2 4 2 2 4

3

5

dx

Q.10 If P = ∫∞

0

x

xdx

2

41+; Q = ∫

∞

0

xdx

x1 4+ and R = ∫

∞

0

dx

x1 4+ then prove that

(a) Q = π4

, (b) P = R, (c) P – 2 Q + R = π

2 2

Q.11 Prove that x n x n a b x nab

x a x bdx

b a

a b

n

a

b n n− − −− + − + +

+ +=

−+z

1 2

2 2

1 12 1

2

b ge j( )( )

( ) ( ) ( )

Q.12 x x

xdx

4 4

2

0

11

1

( )−+∫ Q.13 ∫

1

02

2

x1

xln.x

− dx Q.14 Evaluate: ∫

− +

−2

22

2

dx4x

xx

Q.15 ∫ +−

3

0

2

1 dxx1

x2sin Q.16

0

2π/

∫ ( )xsin

xcosbxsina

4+

+π dx Q.17

0

2π

∫dx

x2 2+ sin

Q.18 ∫π

+π2

0

x dx2

x

4cose Q.19 ∫

−

2

2 2x

1xx12x7x10x3x22

23567

+++−−−+

dx

Q.20 Let α, β be the distinct positive roots of the equation tan x = 2x then evaluate ∫ βα1

0

dx)xsin·x(sin ,

independent of α and β.

Q.21 ∫π 4/

0 x2sin10

xsinxcos

+−

dx Q.22 ∫π

0

( )sec tan

tan

ax b x x

x

++4 2

dx (a,b>0) Q.23 Evaluate: ∫π

++

02

dx)xcos1(

xsin)3x2(

Q.24 If a1, a

2 and a

3 are the three values of a which satisfy the equation

∫ +1

0

3dx)xcosax(sin – ∫−π

1

0

dxxcosx2

a4 = 2

then find the value of 1000( 23

22

21 aaa ++ ).

Q.25 Show that 0

p q+

∫π

| cos x| dx = 2q + sinp where q ε N & − < <π π2 2

p

Q.26 Show that the sum of the two integrals ∫−

−

5

4

e(x+5)² dx + 3 ∫3/2

3/1

e9(x− 2/3)² dx is zero.

Q.27 ∫ +−

−1

02

1

dx1xx

xsinQ.28 ∫

π 2/

0xcosbxsina

xsin2222

2

+dx (a>0, b>0) Q.29 �n

x

x

x

x

1

1 11

1 3

2

+− −−

∫ dx

Q.30 0

2π/

∫ tan−1

−−+

−++

xsin1xsin1

xsin1xsin1dx Q.31 ∫

+

+−−

2

ba

2

ba3

2222

22

22)xb()ax(

xd.x

Q.32 Comment upon the nature of roots of the quadratic equation x2 + 2x = k + ∫ +1

0

dt|kt| depending on the

value of k + R.

Q.33 ∫a2

0

x sin−1

−a

xa2

2

1dx Q.34 Prove that

( )dx

x

dx

xn

nn

1 11

0

1

0 + −= ∫∫

∞

/ (n > 1)

Q.35 Show that ∫∫−− =

x

0

4z4xx

0

zxz dzeedze·e222

Q.36 ∫π

0

( )π−

π

x2

xcos.sin.x2sinx2

2

dx

Q.37 (a) ∫1

0 32 xxx

xd.

x1

x1

+++−

, (b) dxx

1x1n

1xx

1x2

51

124

2

−++−

+∫

+

l

Q.38 Show that dx

x x20 2 1+ +

∞

∫cosθ

= 2 dx

x x20

1

2 1+ +∫cosθ

=

θθ

θ π

θ πθ

θ π π

sin( , )

sin( , )

if

if

∈

−∈

L

N

MMMM

0

22

Q.39 ∫π

+

2

02

2

dxxsin8

xsinx

Q.40 ∫π

+−4

02

2

xdxcos)x2sin1(

)x2cosx2(sinxQ.41 Prove that ∫

x

0

f t dt

u

( )

0

∫

du = ∫x

0

f (u).(x − u) du.

Q.42 ∫π

0

dx

x( cos )5 4 2+Q.43 Evaluate ∫

1

0

ln ( )x1x1 ++− dx Q.44 ∫16

11xtan 1 −− dx

Q.45 ∫

−∞→

+n1

n1

2

ndx|x|)xcos2007xsin2006(nLim . Q.46 Show that f

a

x

x

a

x

xdx a f

a

x

x

a

dx

x( ).

lnln . ( ).

0 0

∞ ∞

∫ ∫+ = +

Q.47 Evaluate the definite integral, ∫− +

++1

1666

6911668998332

dxx1

)xsin·x4xx2(

Q.48 Prove that

(a) α

β

∫ )x()x( −βα− dx = ( )

8

2 πα−β(b)

α

β

∫x

x

−βα− dx = ( )

2

πα−β

(c) α

β

∫)x()x(x

xd

−βα− =

πα β

where α , β > 0 (d) α

β

∫ )x()x(

xd.x

−βα− = ( )2

πβ+α where α < β

Q.49 If f(x) =

4 1 1

1 1 1

1 1

2

2 2 2

2 2 2

cos

(cos ) (cos ) (cos )

(cos ) (cos ) cos

x

x x x

x x x

− + −+ +

, find f x dx( )

−

∫π

π

2

2

Q.50 Evaluate : ∫−−

1

0

1xtann dx)x(cossin·e1l

.

EXERCISE–3Q.1 If the derivative of f(x) wrt x is

)x(f

xcos then show that f(x) is a periodic function .

Q.2 Find the range of the function, f(x) = sin

cos

x dt

t x t1 2 21

1

− +−∫ .

Q.3 A function f is defined in [−1 , 1] as f′(x) = 2 x sin x

1 − cos

x

1; x ≠ 0 ; f(0) = 0;

f (1/π) = 0. Discuss the continuity and derivability of f at x = 0.

Q.4 Let f(x) = [ − − ≤ ≤− < ≤

1 2 0

1 0 2

if x

x if x and g(x) = ∫

−

x

2

f(t) dt. Define g (x) as a function of x and test the

continuity and differentiability of g(x) in (−2, 2).

Q.5 Prove the inequalities:

(a) π6

< dx

x x4

2

82 30

1

− −<∫

π(b) 2 e−1/4 < ∫

−2

0

xx2

e dx < 2e².

dx2

0

<+∫

π

then find a & b. (d) 2

1 ≤ ∫ +

2

02x2

dx ≤

6

5

Q.6 Determine a positive integer n ≤ 5, such that ∫1

0

ex (x − 1)n dx = 16 − 6e.

Q.7 Using calculus

(a) If x < 1 then find the sum of the series ∞++

++

++

++

......x1

x8

x1

x4

x1

x2

x1

18

7

4

3

2.

(b) If x < 1 prove that 284

73

42

3

2 xx1

x21......

xx1

x8x4

xx1

x4x2

xx1

x21

+++

=∞++−

−+

+−−

++−

−.

(c) a < b10 3cos x

(c) Prove the identity f (x)= tanx + 1

2tan

x

2 +

1

22tan

x

22 + .... + 1

2 1n− tanxn2 1− =

1

2 1n− cot xn2 1− – 2cot 2x

Q.8 If φ(x) = cos x − ∫x

0

(x − t) φ(t) dt. Then find the value of φ′′ (x) + φ(x).

Q.9 If y = dt)tx(asin·)t(fa

1x

0

−∫ then prove that yadx

yd 2

2

2

+ = f (x).

Q.10 If y = ∫x

1

tdtnl

x , find xd

yd at x = e.

Q.11 If f(x) = x + ∫1

0

Q.12 A curve C1 is defined by:

dx

dy = ex cos x for x ∈ [0, 2π] and passes through the origin. Prove that the

roots of the function (other than zero) occurs in the ranges 2

π < x < π and

2

3π < x < 2π.

Q.13(a) Let g(x) = xc . e2x & let f(x) = 0

x

∫ e2t . (3 t2 + 1)1/2 dt . For a certain value of 'c', the limit of ′′

f x

g x

( )

( )

as x → ∞ is finite and non zero. Determine the value of 'c' and the limit.

(b) Find the constants 'a' (a > 0) and 'b' such that, 0x

Lim→

t d ta t

bx x

x 2

0+

−

∫

sin = 1.

Q.14 Evaluate: ∫ +−+

+∞→

x3

x

1sin2

2

4

xdt

)3t)(3t(

1t3

dx

dLim

Q.15 Given that Un = {x(1 − x)}n & n ≥ 2 prove that

2n

2

xd

Ud = n (n − 1) U

n−2 − 2 n(2n − 1)U

n−1,

further if Vn = ∫

1

0

ex . Un dx, prove that when n ≥ 2, V

n + 2n (2n − 1).V

n−1− n (n − 1) V

n−2 = 0

Q.16 If ∫∞

0

dttx

tn22 +

�=

π �n 2

4 (x > 0) then show that there can be two integral values of ‘x’ satisfying this

equation.

Q.17 Let f(x) =

1 0 1

0 1 2

2 2 32

− ≤ ≤< ≤

− < ≤

x if x

if x

x if x( )

. Define the function F(x) = ∫x

0

f(t) dt and show that F is

continuous in [0, 3] and differentiable in (0, 3).

Q.18 Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real

′ (0) = 0 & f ′ (1) = 2 ≠ f(0) . Find f(x) & show that, 3 ∫ f(x) dx − x (f(x) + 2) is a constant.

Q.19 Evaluate: (a) ∞→nLim

n/1

2

2

2

2

2

2

2 n

n1.....

n

31

n

21

n

11

+

+

+

+ ;

(b) ∞→nLim 1 1

1

2

2

3

4n n n

n

n++

++ +

..... ; (c) ∞→n

Limn/1

nn

!n

;

(d) Given

n1

nn2

nn3

n C

CLim

∞→=

b

a where a and b are relatively prime, find the value of (a + b).

[xy² + x²y] f(y) dy where x and y are independent variable. Find f(x).

x &

y with f

Q.20 Prove that sin x + sin 3x + sin 5x + .... + sin (2k − 1) x = xsin

xksin 2

, k ∈ N and hence

prove that , ∫π 2/

0 xsin

xksin2

dx = 1k2

1......

7

1

5

1

3

11

−+++++ .

Q.21 If Un= ∫

π 2/

0 xsin

xnsin2

2

dx , then show that U1 , U

2 , U

3 , ..... , U

n constitute an AP .

Hence or otherwise find the value of Un.

Q.22 Solve the equation for y as a function of x, satisfying

x · ∫∫ +=x

0

x

0

dt)t(y·t)1x(dt)t(y , where x > 0, given y (1) = 1.

Q.23 Prove that : (a) Im , n

= ∫1

0

xm . (1 − x)n dx = !)1nm(

!n!m

++ m , n ∈ N.

(b) Im , n

= ∫1

0

xm . (ln x)n dx = (−1)n 1n)1m(

!n++

m , n ∈ N.

Q.24 Find a positive real valued continuously differentiable functions f on the real line such that for all x

f 2(x) = ( ) ( )( )∫ +

x

0

22dt)t(')t( ff + e2

Q.25 Let f(x) be a continuously differentiable function then prove that, ∫x

1

[t] f ′ (t) dt = [x]. f(x) −∑=

]x[

1k

)k(f

where [. ] denotes the greatest integer function and x > 1.

Q.26 Let f be a function such that f(u) − f(v) ≤ u − v for all real u & v in an interval [a, b] . Then:(i) Prove that f is continuous at each point of [a, b] .

(ii) Assume that f is integrable on [a, b]. Prove that, f x dx b a f ca

b

( ) ( ) ( )− −∫ ≤ ( )b a− 2

2, where a ≤ c ≤ b

Q.27 Let F (x) = ∫−

+x

1

2 dtt4 and G (x) = ∫ +1

x

2 dtt4 then compute the value of (FG)' (0) where dash

denotes the derivative.

Q.28 Show that for a continuously thrice differentiable function f(x)

f(x) − f(0) = xf′ (0) + ′′f x( ).0

2

2

+ 1

2

2

0

′′′ −∫ f t x t dtx

( )( )

Q.29 Prove that k

n

=∑

0

(− 1)k ( )kn

1

1k m+ + =

k

m

=∑

0

(− 1)k ( )km

1

1k n+ +

(a) Prove that f (x) + g (x) = 6 for all x. (b) Find f (x) and g (x).

EXERCISE–4

Q.1 Find Limitn → ∞ S

n , if : S

n =

1

2

1

42

1

1

42

4

1

32

2 1n

n n n n

+−

+−

+ ++ −

.......... . [REE ’97, 6]

Q.30 Let f and g be function that are differentiable for all real numbers x and that have the followingproperties:(i) f ' (x) = f (x) – g (x) (ii) g ' (x) = g (x) – f (x)(iii) f (0) = 5 (iv) g (0) = 1

Q.2 (a) If g (x) = cos4

0

t dtx

∫ , then g (x + π) equals :

(A) g (x) + g (π) (B) g (x) − g (π) (C) g (x) g (π) (D) [ g (x)/g (π) ]

(b) Limitn → ∞

1

2 21

2

n

r

n rr

n

+=∑ equals :

(A) 1 5+ (B) − +1 5 (C) − +1 2 (D) 1 2+

(c) The value of π πsin ( ln )x

x

e

1

37

∫ dx is _______ .

(d) Let d

dxF x

e

x

x

( )sin

= , x > 0 . If 2

2

1

4e

x

xsin

∫ dx = F (k) − F (1) then one of the possible

values of k is ______.

(e) Determine the value of 2 1

12

x x

xdx

( sin )

cos

+

+−∫π

π. [JEE ’97, 2 + 2 + 2 + 2 + 5]

Q.3 (a) If f t dt x t f t dtx

x

( ) ( )= + ∫∫1

0

, then the value of f (1) is

(A) 1/ 2 (B) 0 (C) 1 (D) − 1/ 2

(b) Prove that tan tan− −

− +

=∫ ∫

1

20

11

0

11

12

x xdx x dx . Hence or otherwise, evaluate the integral

( )tan− − +∫1 2

0

1

1 x x dx [JEE’98, 2 + 8]

Q.4 Evaluate 1

5 2 2 12 2 40

1

( )( )( )+ − + −∫x x e

dxx

[REE ’98, 6 ]

Q.5 (a) If for al real number y, [y] is the greatest integer less than or equal to y, then the value of the

integral π

π

/

/

2

3 2

∫ [2 sin x] dx is :

(A) − π (B) 0 (C) − π2

(D) π2

(b) dx

x14

3 4

+∫cos

/

/

π

π is equal to :

(A) 2 (B) − 2 (C) 1

2(D) −

1

2

(c) Integrate :

( )x x

x x

3

22

3 2

1 1

+ +

+ +∫

( ) dx

(d) Integrate: ∫π

−+0

xcosxcos

xcos

dxee

e[JEE '99, 2 + 2 + 7 + 3 (out of 200)]

Q.6 Evaluate the integral 3 2 1

0

6 cos

cos

/ x

x

−∫

π

dx. [ REE '99, 6]

Q.7 (a) The value of the integral e

e

−∫

1

2

loge x

x d x is :

(A) 3/2 (B) 5/2 (C) 3 (D) 5

(b) Let g (x) = 0

x

∫ f (t) d t , where f is such that 1

2 ≤ f (t) ≤ 1 for t ∈ (0, 1] and 0 ≤ f (t) ≤

1

2for t ∈ (1, 2]. Then g (2) satisfies the inequality :

(A) − 3

2 ≤ g (2) <

1

2(B) 0 ≤ g (2) < 2 (C)

3

2 < g (2) ≤

5

2(D) 2 < g (2) < 4

(c) If f (x) = { e x for x

otherwise

xcos. sin | | ≤ 2

2. Then

−∫2

3

f (x)d x :

(A) 0 (B) 1 (C) 2 (D) 3

(d) For x > 0, let f (x) = 1

x

∫�n t

t1 + dt. Find the function f (x) + f (1/x) and show that,

f (e) + f (1/e) = 1/2 . [JEE 2000, 1 + 1 + 1 + 5]

Q.8 (a) Sn =

1

1 + n +

1

2 2+ n + ........ +

1

2n n+ . Find Limit

n → ∞ Sn .

(b) Given 0

1

∫sin t

t1 + d t = α , find the value of

4 2

4

π

π

−∫

sin t

t

2

4 2π + − d t in terms of α .

[ REE 2000, Mains, 3 + 3 out of 100]

Q.9 Evaluate sin− +

+ +

∫

1

2

2 2

4 8 13

x

x xdx .

Q.10 (a) Evaluate cos

cos sin

/ 9

3 30

2x

x xdx

+∫π

. (b) Evaluate xdx

x1+∫cos sinα

π

0

[ REE 2001, 3 + 5]

Q.11 (a) Let f(x) = 2 2

1

−∫ t dt

x

. Then the real roots of the equation x2 – f ′ (x) = 0 are

(A) +1 (B) + 1

2(C) +

1

2(D) 0 and 1

(b) Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R

f (x + T) = f (x). If I = f( )x

T

0

∫ dx then the value of f(2 )x

T

3

3 3+

∫ dx is

(A) 3

2 I (B) 2 I (C) 3 I (D) 6 I

(c) The integral [ ]x nx

x+

+−

−

∫ l1

11

2

1

2

dx equals

(A) – 1

2(B) 0 (C) 1 (D) 2ln

1

2

[JEE 2002(Scr.), 3+3+3](d) For any natural number m, evaluate

x x x x x dxm m m m m m3 2 2

1

2 3 6+ + + +z c h c h , where x > 0 [JEE 2002 (Mains),4]

Q.12 If f is an even function then prove that ∫π 2

0

f (cos2x) cosx dx = ∫π 4

0

2 f (sin2x) cosx dx

[JEE 2003,(Mains) 2 out of 60]

Q.13 (a) ∫ +−1

0

dxx1

x1 =

(A) 12

+π

(B) 12

−π

(C) π (D) 1

(b) If 5t

0

t5

2dx)x(fx

2

=∫ , t > 0, then

25

4f =

(A) 5

2(B)

2

5(C) –

5

2(D) 1

[JEE 2004, (Scr.)]

(c) If ( ) θθ+θ

= ∫π

d.sin1

cos.xcosxy

2

16/2

x

2 then find

dx

dy at x = π. [JEE 2004 (Mains), 2]

(d) Evaluate dx

3|x|cos2

x43/

3/

3

∫π

π−

π+−

+π. [JEE 2004 (Mains), 4]

Q.14 (a) If ( )∫1

xsin

2 dt)t(ft = (1 – sin x), then f

3

1 is [JEE 2005 (Scr.)]

(A) 1/3 (B) 31 (C) 3 (D) 3

(b) ( )∫−

++++++0

2

23 dx)1xcos()1x(3x3x3x is equal to [JEE 2005 (Scr.)]

(A) – 4 (B) 0 (C) 4 (D) 6

(c) Evaluate: ∫π

+

0

|xcos| dxxsinxcos2

1cos3xcos

2

1sin2e . [JEE 2005, Mains,2]

Q.15 dx1x2x2x

1x

243

2

∫+−

− is equal to

(A) 2

24

x

1x2x2 +− + C (B)

3

24

x

1x2x2 +− + C

(C) x

1x2x2 24 +− + C (D)

2

24

x2

1x2x2 +− + C [JEE 2006, 3]

Comprehension

Q.16 Suppose we define the definite integral using the following formula ( ) ( ))b(f)a(f2

abdxxf

b

a

+−

=∫ , for

more accurate result for c ∈ (a, b) F(c) = ( ) ( ))c(f)b(f2

cb)c(f)a(f

2

ac+

−++

−. When c =

2

ba +,

( ) ))c(f2)b(f)a(f(4

abdxxf

b

a

++−

=∫

(a) ∫π 2/

0

dxxsin is equal to

(A) ( )218

+π

(B) ( )214

+π

(C) 28

π(D)

24

π

(b) If f (x) is a polynomial and if

( )

( )0

at

)a(f)t(f2

atdx)x(f

Lim3

t

a

at=

−

+−

−∫

→ for all a then the degree of f (x) can

atmost be(A) 1 (B) 2 (C) 3 (D) 4

(c) If f ''(x) < 0, ∀ x ∈ (a, b) and c is a point such that a < c < b, and (c, f (c) ) is the point lying on the curvefor which F(c) is maximum, then f '(c) is equal to

(A) ( ) ( )

ab

afbf

−−

(B) ( ) ( )( )

ab

afbf2

−−

(C) ( ) ( )

ab2

afbf2

−−

(D) 0

[JEE 2006, 5 marks each]

Q.17 Find the value of

( )

( )∫

∫

−

−

1

0

10150

1

0

10050

dxx1

dxx15050

[JEE 2006, 6]

ANSWER

EXERCISE–1

Q.1 ln

θθ++

2cos

2cos311 2

+ C Q.2 − x

x x

++ +

1

15 + c

Q.3 1

4ln(cos x + sin x) +

x

2 +

8

1 (sin 2x + cos 2x) + c Q.4

3

8 tan−1 x − ( )

x

x4 14 −−

3

16 ln

x

x

−+

1

1 + c

Q.5 2 tan−1 x x x+ + −

2 2 1 + c Q.6 x

e

e

xc

x x −

+

Q.7 (c) 2

1 (sin 2 θ) ln cos sin

cos sin

θ θθ θ

+−

−

2

1 ln (sec 2 θ) + c Q.8

2

1 ln

2xtan

xtan

+ + c

Q.9

+

+−

2

21

22 b

xtanatanx

ba

1+ c Q.10 2ln

1t2

t

+ +

1t2

1

+ + C when t = x + xx2 +

Q.11 3

1

2/32

2xx

++ − 2/1

2 2xx

2

++

+ c

Q.12 cos a . arc cos

acos

xcos − sin a . ln

−+ asinxsinxsin 22

+ c Q.13 ( )

3/8

2

)x(tan8

xtan413 +− + c

Q.14 2

1 ln

2

xtan +

4

1 sec²

2

x + tan

2

x + c Q.15 cxcosarcx12x1x ++−−−

Q.16 (a + x) arc tan a

x − xa + c Q.17

( )

+−++

23

22

x

11ln32.

x9

1x1x

Q.18 xln (lnx) − xnl

x + c Q.19 xx

x

ex1

1

ex1

exln

++

+ + c

Q.20 − ln (1 − x4) + c Q.21

−+++− − ttan)t1(n

2

1t

2

t

4

t6 12

24

l + C where t = x1/6

Q.22 4

2cos x

+ 2 tan−1 cos x2 − ln

1

1

2

2

+

−

cos

cos

x

x+ c Q.23 C – ln(1 + (x + 1)e–x) – xe)1x(1

1−++

Q24. sin−1

2

xsec

2

1 2 + c

Q.25 c)xcos3xsin34(

)xcos3xsin34(nl

24

1+

−−++

Q.26 2

1

π+−−

82

xtannl

2

1xcosxsin + c

Q.27 c)xcosx(sintanarcxcosxsin3

xcosxsin3nl

32

1+++

+−−+ Q.28 − − +

� � �n x n x n x(sec ) (sec ) (sec )

1

22

1

33 + c

Q.29 − 1

sinα ln [ ]cot cot cot cot cotx x x+ + + −α α2 2 1 + c Q.30 ln

xsinxcosx

xcosxsinx

−+

Q.31 c12

xtantanarc3x2 +

+− Q.32 cos

sin

2x

x − x − cot x . ln ( )( )e x xcos cos+ 2 + c

Q.33 ln (1 + t) − 1

4 ln (1 + t4) +

1

2 2 ln

t t

t t

2

2

2 1

2 1

− ++ +

− 1

2 tan−1 t2 + c where t = cotx

Q.34 c2

xtan

4

1

2

xtannl

2

1 2 +− Q.35 22 )1x(

xc

−−

Q.36 c – ecos x (x + cosec x) Q.37 sin− +

+1

2ax b

cxk Q.38 ex

x1

x1

−+

+ c Q.39 cx10x79

)20x7(2

2+

−−

−

Q.40 c1x

xnlxsecarc

2+

−− Q.41 �n

u

u u

uc where u

x

x

| |tan

2

4 2

12

31

13

1 2

3

1

1

−

+ ++

++ =

−+

−

Q.42 ( )8

3

1

2 5

5 1

5 11

1 1 2tan sin

− −+−+

− − −t nt

tx x� + c where t =

1

1

+−

x

x

Q.43 tan−1 2 2sin

sin cos

x

x x+

+ c Q.44 4 ln x +

x

7 + 6 tan–1(x) + 2x1

x6

+ + C

Q.45 c)1x(3

xtanarc

3

2+

+Q.46 −

− −+

− + − −

−+

−2 2

4

4 2 2 2 2 1

3

2 21x x

xn

x x x

x

x� sin + c

Q.47 cx

x.

2+

α−β−

β−α−

Q.48 2

1 ln

−

+++++ 122x

1x2

x

1x

2

+ C

Q.49

+−

−

−+

t1

t1n

2

1

t2

t2n

2

1ll where t = cosθ and θ = cosec–1(cotx)

Q.50

α

−

α −

2eccos

x2

1xtan·

2eccos

2

1 21

EXERCISE–2

Q.1 6

2π Q.2 2nl Q.3 6 − 2e Q.4

π2

1− Q.5 5

64

πQ.6

8

π ln 2

Q.7 1 – sec(1) Q.8 62 Q.9 2 2 + 4

3 ( )3 3 2 2− Q.12

22

7−

π Q.13

8

π(1 − ln 4)

Q.14 )12(n424 +− l Q.15 3

3π Q.16

22

)ba( +π Q.17

2

3

πQ.18 – )1e(

5

23 2 +π

Q.19 5

216

22−

πQ.20 0 Q.21

−

3

1tanarc

3

2tanarc

3

1

Q.22 ( )a bπ π+ 2

3 3Q.23

2

)3( +ππQ.24 5250 Q.27

36

2π

Q.28 )ba(a2 +

πQ.29

5

3

πQ.30

16

3 2πQ.31

12

πQ.32 real & distinct ∀ k ∈ R

Q.33 4

a2πQ.36

π8

Q.37 (a) 3

π; (b) 2n

8l

π Q.39 –

3

2 2π ln 2 Q.40

π2

16 −

π4

ln2

Q.42 27

5π Q.43

1

22

21ln + −

πQ.44 32

3

16−

πQ.45 2007 Q.47

666

4+π

Q.49 –2π – 32

15Q.50

2

1)2n1(

48

2

++π

−π

l

EXERCISE–3

Q.2 −

π π2 2

, Q.3 cont. & der. at x = 0

Q.4 g(x) is cont. in (−2 , 2); g(x) is der. at x = 1 & not der. at x = 0 . Note that ;

g(x) =

− + − ≤ ≤− + − < <

− − ≤ ≤

( )x for x

x for x

x for x

x

x

2 2 0

2 0 1

1 1 2

2

22

2

Q.5 (c) a = 13

2π & b =

7

2π Q.6 n = 3

Q.7 (a) x1

1

−Q.8 – cos x Q.10 1 + e Q.11 f(x) = x +

119

61 x +

119

80 x²

Q.13 (a) c = 1 and Limitx → ∞ will be

3

2 (b) a = 4 and b =1 Q.14 13.5

Q.16 x = 2 or 4 Q.17 F(x) =

( )3x2if

2x1if

1x0ifx

21

3

2x

21

2x

3

2

≤<+

≤<≤≤−

−

Q.18 f (x) = 1 + x2 Q.19 (a) 2 e(1/2) (π − 4); (b) 3 − ln 4; (c) e

1; (d) 43 Q.21 U

n =

2

nπ

Q.22 y = x1

3e

x

e −Q.24 f (x) = ex + 1 Q.27 0

Q.30 f (x) = 3 + 2e2x; g (x) = 3 – 2e2x

EXERCISE–4Q.1 π/6 Q.2 (a) A (b) B (c) 2 (d) 16 (e) π² Q.3 (a) A (b) ln2

Q.4 1

2 11

11 1

11 1�n

+−

Q.5 ( a )( a )( a )( a ) C, (b) A ; (c) 3

2 tan-1 x −

1

2 ln (1 + x) +

1

4 ln (1 + x2) +

x

x1 2+ + c, (d)

2

π

Q.6 2

3 π − 2 tan−1 2 Q.7 (a) B, (b) B, (c) C, (d)

1

2 ln2 x

Q.8 (a) 2 ln 2, (b) – α Q.9 (x + 1) tan–1 )13x8x4(n4

3

3

)1x(2 2 ++−+

� + C

Q.10 (a) 1

8

5

4

1

3

π−

, (b) I =

ππ∈απ−αα

π

π∈αα

πα

)2,(if)2(sin

),0(ifsin

Q.11 (a) A, (b) C, (c) B, (d) 1

6 12 3 63 2

1

( )mx x x Cm m m

m

m

++ + +

+

d i

Q.13 (a) B, (b) A, (c) 2π, (d)

π −

2

1tan

3

4 1

Q.14 (a) C, (b) C, (c)

−

+

1

2

1sin

2

e

2

1cose

5

24

Q.15 D Q.16 (a) A, (b) A, (c) A Q.17 5051

ELEMENTARY DEFINITE INTEGRAL

(SELF PRACTICE)Evaluate the following definite integrals.

Q.1 sin

( )

−

−∫1

0

1

1

x

x x

dx Q.2 0

2�n

∫ x e−x dx Q.3 sin

cos

/x dx

x1 20

3 4

+∫π

Q 4. 0

2π /

∫ e2x . cos x dx Q 5. x dx

x5 41

1

−−∫ Q 6.

1 12

2� �n x n x

e

−

∫ dx

Q 7. sin

sin cos

/2

4 40

4x

x x+∫π

dx Q 8. cos

( sin ) ( sin )

/x dx

x x1 20

2

+ +∫π

Q 9. ( )

sin . cos

sin cos

/ 2 2

3 32

0

4x x

x x+∫

π dx

Q 10. ( ) ( )x x− −∫ 1 21

2

dx Q 11. dx

x x( ) ( )− −∫1 52

3

Q 12. dx

x x( )

/

+ +∫

1 1 20

3 4

Q 13. ( )sin cos sin cos/

φ φ φ φπ

a b2 2 2 2

0

2

+∫ dφ a≠b Q14. x x2 2

0

1

4. −∫ dx

Q 15. 0

4π /

∫ x cos x cos 3x dx Q16. dx

x5 40

2

+∫sin

/π

Q17. dx

x x x( )− −∫

1 222

3

Q 18. dx

x10

2

+∫cos . cos

/

θ

π

θ ∈ (0, π) Q 19. x dx

x x+ + +∫1 5 10

3

Q 20. ( )

dx

x12

3 2

1

3

+∫ /

Q 21. 0

2π /

∫ sin4x dx Q 22. 0

4π /

∫ cos 2x 1 2− sin x dx Q23. x

x30

3

−∫ dx

Q.24 ( )dx

x x1 2 12 20

1 2

− −∫/

Q 25. ( )dx

x x41

2

1+∫ Q 26. xa x

a x

a 2 2

2 20

−+∫ dx

Q27. sin cos

cos cos

/x x

x x2

0

2

2 2+ +∫π

dx Q 28. 0

1

∫ x (tan−1 x)2 dx Q 29.

( )sin

/

−

−∫

1

23 2

0

12

1

x

x

dx

Q 30. dx

x x20

1

2 1+ +∫cosα

where −π < α < π Q 31. x

x

2

40 1+

∞

∫ dx

Q 32. dx

xa

b

1 2+∫ where a =

e e− −1

2 & b =

e e2 2

2

− −

Q33. 1

1

2

2 4

0

1 −+ +∫

x

x xdx

Q 34. xx

x

52

20

11

1∫

dx Q 35. dx

x x3 20

+ +∫sin cos

π

Q 36. sin cos

sin

/ θ θθ

π ++∫

9 16 20

4

dθ

Q 37. 0

π

∫ θ sin2 θ cos θ dθ Q 38. 1 2

22

0

2 ++∫

cos

( cos )

/x

x

π

dx Q 39. x x

x

++∫

sin

cos

/

10

2π

dx

Q 40. 0

2π /

∫ cos3x sin 3x dx Q 41. 2

1 1

2

20

1 −

+ −∫

x

x x( )

dx Q 42. dxe1

1

dx

d1

1

x/1∫−

+

+−

Q 43. ∫e

0xx )exln(

dxQ 44.

−∫1

1

x2 d (ln x)

Q 45. If f(π) = 2 & 0

π

∫ (f(x) + f ′′ (x)) sin x dx = 5, then find f(0)

Q.46 ∫b

a

dxx

|x|Q.47 cos cos

2 2

0

3

8 4

11

8 4

π ππ

−

− +

∫

x x dx

Q.48sec tan

sec tan

cos

cos

/x x

x x

ecx

ecx

−+ +∫

1 20

2π

dx Q.49 dx)x(''fx

1

0

∫ , where f (x) =cos(tan–1x)

Q.50�

�

n

n

2

3

∫ f (x)dx, where f(x) = e −x + 2 e −2x + 3 e −3x + ......∞

ANSWER KEY

Q 1. π2

4Q 2.

1

2 2�n

e Q3. π

4

1

2

1+ −tan Q 4. e

π − 2

5

Q 5. 1

6Q6. e −

2

2�nQ 7.

π4

Q 8. ln 4

3

Q 9. 1

6Q 10.

π8

Q 11. π6

Q 12. 1

2

9 4 2

7�n

+

Q 13. 1

3

a b

a b

3 3

2 2

−−

Q 14. π3

3

4− Q 15.

π − 3

16Q 16.

2

3 tan−1

1

3

Q 17. π3

Q 18 θ

θsinQ 19.

14

15Q 20.

3 2

2

−

Q 21. 3

16

πQ 22.

1

3Q 23.

3

2

πQ 24. ( )1

22 3�n +

Q 25. 1

4 ln

32

17Q 26.

a2

4 (π − 2) Q 27.

π4

− tan−1 2 + 1

2 ln

5

2

Q 28. π π4 4

11

22−

+ �n Q 29.

π4

1

2− ln 2 Q 30.

αα

α α2

01

20

sin;if if≠ =

Q 31. π

2 2Q 32. 1 Q 33.

1

2 ln 3 Q 34.

3 8

24

π +

Q 35. π4

Q 36. 1

20 ln 3 Q 37. −

4

9Q 38.

1

2

Q 39. π2

Q 40. 5

12Q 41.

π2

Q 42. e1

2

+

Q 43. ln 2 Q 44. e e2 2

2

− −

Q 45. 3 Q.46 | b | – | a |

Q.47 2 Q.48 π/3 Q.49 22

31− Q.50

2

1

EXERCISE–5Part : (A) Only one correct option

1. ∫ ′′−′′ )]x(g)x(f)x(g)x(f[ dx is equal to

(A) )x(g

)x(f

′ (B) f′(x) g(x) – f(x) g′(x)

(C) f(x) g′(x) – f′(x) g(x) (D) f(x) g′(x) + f′(x) g′(x)

2. ∫xcosxsin

1

3 dx is equal to

(A) xtan

2− + c (B) xtan2 + c (C)

xtan

2 + c (D) – xtan2 – c

3. ∫�

�

n x

x n x

| |

| |1 + dx equals :

(A)2

31 + �n x (�nx − 2) + c (B)

2

31 + �n x (�nx + 2) + c

(C)1

31 + �n x (�nx − 2) + c (D) 2 1 + �n x (3 �nx − 2) + c

4. If ∫+

−

2

1

x1

xtanx dx = 2x1+ f(x) + A A �n (x + 1x2 + ) + C, then

(A) f(x) = tan–1 x, A = –1 (B) f(x) = tan–1 x, A = 1(C) f(x) = 2 tan–1 x , A = –1 (D) f(x) = 2 tan–1 x, A = 1

5. ∫ −

−

xcosxsin21

xcosxsin22

88

dx =

(A) 2

1 sin 2x + c (B) –

2

1 sin 2x + c (C) –

2

1 sin x + c (D) – sin2x + c

6. ∫ +−

−−+

xa

xa

xa

xa dx is equal to

(A) – 2 22 xa − + C (B) 22 xa − + C (C) – 22 ax − + C (D) none of these

7. ∫ α+α− )xtan()xtan( tan 2x dx is equal to

(A) �n )xsec(

)xsec(.x2sec

α−α+

+ C (B) �n )xsec()xsec(

x2sec

α+α− + C

(C) �n )xsec(

)xsec(.x2sec

α+α+

+ C (D) none of these

8. ∫ −1xsec dx is equal to

(A) 2 �n

−+

2

1

2

xcos

2

xcos 2

+ C (B) �n

−+

2

1

2

xcos

2

xcos 2

+ C

(C) – 2 �n

−+

2

1

2

xcos

2

xcos 2


9. ∫ x2sinxcos

dx3 is equal to

(A) 2

+ xtan5

1xcos 2/5

+ C (B) 2

+ xtan5

1xtan 2/5

+ C

(C) 2

− xtan5

1xtan 2/5


10. Primitive of

( )3 1

1

4

4 2

x

x x

−

+ + w.r.t. x is:

(A)x

x x4 1+ + + c (B) −

x

x x4 1+ + + c

(C)x

x x

++ +

1

14 + c (D) −x

x x

++ +

1

14 + c

11. If( )x

x x

4

2 2

1

1

+

+∫ dx = A �n x +

B

x1 2+ + c, where c is the constant of integration then:

(A) A = 1; B = − 1 (B) A = − 1; B = 1(C) A = 1; B = 1 (D) A = − 1; B = − 1

12. ∫1

1

−+

x

x dx equals :

(A) x 1 − x − 2 1 − x + cos −1 ( )x + c (B) x 1 − x + 2 1 − x + cos −1 ( )x + c

(C) x 1 − x − 2 1 − x − cos −1 ( )x + c (D) x 1 − x + 2 1 − x − cos −1 ( )x + c

13. ∫ sin x. cos x. cos 2x. cos 4x. cos 8x. cos 16 x dx equals:

(A)sin 16

1024

x + c (B) −

cos 32

1024

x + c (C)

cos 32

1096

x + c (D) −

cos 32

1096

x + c

14. ∫1

6 6cos sinx x+ d

x equals :

(A) tan −1 (tan x + cot x) + c (B) − tan −1 (tan x + cot x) + c(C) tan −1 (tan x − cot x) + c (D) − tan −1 (tan x − cot x) + c

15. ∫

−π

++ dx2

x

4tanx)xsin1ln( is equal to:

(A) x �n (1 + sinx) + c (B) �n (1 + sin x) + c (C) – x �n (1 + sin x) + c (D) �n (1 – sin) + c

16.dx

x xcos . sin3

2∫ equals:

(A)2

5 (tan x)5/2 + 2 tanx + c (B)

2

5 (tan2 x + 5) tanx + c

(C)2

5 (tan2 x + 5) 2tanx + c (D) none

17. Ifdx

x xsin cos3 5∫ = a cot x + b tan3 x + c where c is an arbitrary constant of integration then the

values of ‘a’ and ‘b’ are respectively:

(A) − 2 & 2

3(B) 2 & −

2

3(C) 2 &

2

3(D) none

18. ∫+−

−

1x2x2x

1x

243

2

dx is equal to [IIT - 2006, (3, –1)]

(A) 2

24

x

1x2x2 +− + c (B)

3

24

x

1x2x2 +− + c

(C) x

1x2x2 24 +− + c (D)

2

24

x2

1x2x2 +− + c

Part : (B) May have more than one options correct

19. If( )x dx

x x x

−

− +∫

1

2 2 12 2 is equal to

f x

g x

( )

( ) + c then

(A) f(x) = 2x2 – 2x + 1 (B) g(x) = x + 1

(C) g(x) = x (D) f(x) = x2x2 2 −

20.dx

x5 4+∫cos

= Ι tan−1 mx

tan2

+ C then:

(A) l = 2/3 (B) m = 1/3 (C) l = 1/3 (D) m = 2/3

21. If3 3

3 3

cot cot

tan tan

x x

x x

−−∫ dx = p f(x) + q g(x) + c where 'c' is a constant of integration, then

(A) p = 1; q =1

3; f(x) = x; g(x) = �n

3

3

−+

tan

tan

x

x

(B) p = 1; q = −1

3; f(x) = x; g(x) = �n

3

3

−+

tan

tan

x

x

(C) p = 1; q = −2

3; f(x) = x; g(x) = �n

3

3

+−

tan

tan

x

x

(D) p = 1; q = −1

3; f(x) = x; g(x) = �n

3

3

+−

tan

tan

x

x

22.sin

sin cos

24 4

x

x x+∫ dx is equal to:

(A) cot −1 (cot2 x) + c (B) − cot −1 (tan2 x) + c(C) tan −1 (tan2 x) + c (D) − tan −1 (cos 2 x) + c

23.

�n

x

xx

−+

−∫

11

21

dx equal:

(A)1

2 �n2

x

x

−+

1

1 + c (B)

1

4 �n2

x

x

−+

1

1 + c (C)

1

2 �n2

x

x

+−

1

1 + c (D)

1

4 �n2

x

x

+−

1

1 + c

24.�n x

x x

(tan )

sin cos∫ dx equal:

(A)1

2 �n2 (cot x) + c (B)

1

2 �n2 (sec x) + c

(C)1

2 �n2 (sin x sec x) + c (D)

1

2 �n2 (cos x cosec x) + c

EXERCISE–6

1. Integrate with respect ∫ − )xcosx(sin

xsin.xeccos 2

. dx

2. Integrate with respect to x 42

2

xx1

x1

+−−

3. Integrate with respect to x 2x)1x(

1

++ 2

4. ∫( )x

x x

−

+ +

1

1

2

4 2 dx 5. ∫

2 2

6 42

sin cos

cos sin

φ φ

φ φ

−

− − dφ

6. ∫tan tan

tan

θ θ

θ

+

+

3

31 dθ 7. ∫

cos cos

cos

5 4

1 2 3

x x

xdx

+−

8. ∫3 4 2

3 2

+ ++ +

sin cos

sin cos

x x

x x dx 9. ∫

1 ++

cos cos

cos cos

αα

x

x dx

10. ∫d x

x x x( ) ( ) ( )− − −α α β11. ( )

dx

x x x x x3 2 23 3 1 2 3+ + + + −∫

12. ∫ ex

( )x x

x

3

22

2

1

− +

+ dx 13. ∫

−

−dx

xcot4xcos

)3x2(cos

24

14. ∫( ){ }

−++4

22

x

xln21xnl1x dx 15. ∫ ( )

d x

a b x+ cos2

, (a > b)

16. ∫cos cot

cos cot.

sec

sec

ec x x

ec x x

x

x

−+ +1 2

dx 17. ∫ ( )x

x x7 102

3 2

− −/ dx

18. ∫2 2

2

− −x x

x dx 19. ∫ tan −1 x. �n (1 + x2) dx.

20.( )

a b x

b a x

+

+∫

sin

sin2 dx 21.

( )dx

x x4 32

1+∫

22. ( )1

1 2 2

+−∫

x x

x x e x

cossin

dx

23.

( )x

x x

cos

cos/

α

α

+

+ +∫

1

2 123 2

dx =f x

g x

( )

( ) + c then find f(x) and g(x)

24. Evaluate ∫+

xcos

)xsin1(n2

2�

dx.

25. Integrate,

( )x x

x x

3

22

3 2

1 1

+ +

+ +∫

( ) dx. [IIT - 1999, 7]

26. For any natural number m, evaluate,

∫ ( )x x xm m m3 2+ + ( )2 3 62

1x x

m mm

+ +/

d x, x > 0. [IIT - 2002, 5]

ANSWER

EXERCISE–51. C 2. A 3. A 4. A 5. B 6. A

7. B 8. C 9. B 10. B 11. C 12. A

13. B 14. C 15. A 16. B 17. A 18. D

19. AC 20. AB 21. AD 22. ABCD 23. BD

24. ACD

EXERCISE–6

1. �n 2

xtan21+ + c 2. –

32

1 �n

xx

1x

3x

1x

++

−+ + c

3. – 3

1 �n 9

2

3

1t

3

1t

2

+

−+

− + c

where t= 1x

1

+

4.1

3 tan −1

−

3x

1x2

−3

2 tan −1

+

3

1x2 2

+ c

5. 2 �n 5sin4sin2 +φ−φ + 7 tan −1(sinφ− 2) + c

6. − 1

3�n1+tanθ+

1

6�ntan2 θ − tan θ + 1+

1

3

tan− 12 1

3

tan θ −

+ c

7. − +(sinsin

)xx2

2+ c

8. 2 32

1x arcx

c− +

+tan tan

9. x cos α + sin α �ncos ( )

cos ( )

1212

α

α

−

+

x

x + c

10. cx

x.

2+

α−β−

β−α−

11.x x

x

2

2

2 3

8 1

+ −+( )

+1

16. cos−1

2

1x +

+ c

12. exx

x

+

+

1

12 + c

13. c – 3

1 tanx.(2 + tan2x). xcot4 2−

14.( )x x

x x

2 2

3 2

1 1

92 3 1

1+ +− +

. ln

15. – ( ) )xcosba(ba

xsinb22 +−

+ ( ) 2/322 ba

a2

−

arctanba

ba

+−

.tan2

x + c

16. sin−11

2 2

2sec

x

+ c

17.2 7 20

9 7 102

( )x

x xc

−

− −+

18. –x

xx2 2−−+

4

2ln

−−+−x

xx222x4 2

– sin–1

+3

1x2+ c

19. x tan −1 x. �n (1 + x2) + (tan −1 x)2 − 2x tan −1 x

+ �n (1 + x2) − �n x1 22

+

+ c

20. – cxsinab

xcos+

+

21.2

3 �n 3

3

x

1x + −

1

3 3x − ( )

1

3 13x + + c

22. ln (x esinx) -1

2ln (1 - x2 e2 sinx) + c

23. x; x2 + 2x cos α + 1

24. tan x ln (1 + sin2x) – 2x + 2 tan–1 ( 2 .tan x) + c.

25.3

2 tan-1 x -

1

2 �n (1 + x) +

1

4 �n (1 + x2) +

x

x1 2+ + c

26.z

m

m

m

+

+

1

6 1( ) + c, where z = 2

x3m + 3

x2m + 6

xm

EXERCISE–7Part : (A) Only one correct option

1. If f(x) is a function satisfying f

x

1+ x2 f(x) = 0 for all non-zero x, then ∫

θ

θ

eccos

sin

dx)x(f equals

(A) sinθ + cosecθ (B) sin2 θ (C) cosec2 θ (D) none of these

2. The value of the integral ∫ +α+

1

0

2 1cosx2x

dx, where 0 < α <

2

π, is equal to

(A) sin α (B) α sin α (C) α

αsin2

(D) 2

αsin α

3. If ∫100

0

)x(f dx = a, then ( )∑ ∫=

+−

100

1r

1

0

dxx1rf =

(A) 100 a (B) a (C) 0 (D) 10 a

4. If f(x) is an odd function defined on

−2

T,

2

T and has period T, then φ(x) = ∫

x

a

dt)t(f is

(A) a periodic function with period 2

T(B) a periodic function with period T

(C) not a periodic function (D) a periodic function with period 4

T

5. If f(π) = 2 and ∫π

′′+0

))x(f)x(f( sin x dx = 5 then f(0) is equal to, (it is given that f(x) is continuous in [0, π])

(A) 7 (B) 3 (C) 5 (D) 1

6. If f(0) = 1, f(2) = 3, f′(2) = 5 and f′(0) is finite, then ∫ ′′1

0

f.x (2x) dx is equal to

(A) zero (B) 1 (C) 2 (D) none of these

7. ∞→nlim

n/1

n

)1n(sin.......

n2

3sin.

n2

2sin.

n2sin

π−πππ is equal to

(A) 2

π(B) e4/π (C) e2/π (D) none of these

8. f(x) = Minimum {tanx, cot x} ∀ x ∈

π2

,0 . Then ∫π 3/

0

)x(f dx is equal to

(A) �n

2

3(B) �n

2

3(C) �n ( 2 ) (D) �n ( 3 )

9. If A = ∫π

+0

2)2x(

xcos dx, then ∫

π

+

2

01x

x2sin dx is equal to

(A) 2

1 +

2

1

+π – AA (B)

2

1

+π – AA (C) 1 +

2

1

+π – AA (D) A –

2

1 –

2

1

+π

10. ∫π

π−+

2/

2/

2 1x2cos8

dx|x| has the value

(A) 6

2π(B)

12

2π(C)

24

2π(D) none of these

11. ∞→nLt ∑

+= −

n3

1n2r22 nr

n is equal to

(A) log 3

2(B) log

2

3(C) log

3

2(D) log

2

3

12. If ∫y

a

2dttcos = ∫2x

at

tsindt , then the value of

dx

dy is

(A) ycosx

xsin22

2

(B) 2

2

ycosx

xsin2(C)

−

2

ysin21x

xsin22

2

(D) none of these

13. If f(x) =

=

+=

whereelse,1

.....3,2,1n,1n

nxwhere,0

, then the value of ∫2

0

dx)x(f

(A) 1 (B) 0 (C) 2 (D) ∞

14. ∫ −

1

0

4/3)x1(

dxx =

(A) 16

15(B) –

5

16(C) –

16

3(D) none

15. Let Ι1 = ∫

+

2

12x1

dx and Ι

2 = ∫

2

1x

dx, then

(A) Ι1 > Ι

2(B) Ι

2 > Ι

1(C) Ι

1 = Ι

2(D) Ι

1 > 2Ι

2

16. The value of ∫ −]x[

0

])x[x( dx is

(A) 2

1[x] (B) 2[x] (C)

]x[2

1(D) none of these

17. The value of the integeral ∫−

−−

++

+

3

1

21

2

1

x

1xtan

1x

xtan dx is equal to

(A) π (B) 2π (C) 4π (D) none of these

18. The value of ∫π

+2/

0

|xcotxtan|log dx is

(A) π log 2 (B) –π log 2 (C) 2

π log 2 (D) –

2

π log 2

19. If ∫1

0

x2

e (x – α) dx = 0, then

(A) 1 < α < 2 (B) α < 0 (C) 0 < α < 1 (D) α = 0

20. Suppose for every integer n, ∫+

=1n

n

2ndx)x(f . The value of ∫−

4

2

dx)x(f is :

(A) 16 (B) 14 (C) 19 (D) 21

21. Let A = 0

1

∫e d t

t

t

1 + dt then

e

t a

t

a

a −

− − −∫1

1

dt has the value :

(A) Ae–a (B) – Ae–a (C) – ae–a (D) Aea

22.( )

∫−

5

2/5

4

32

x

x25 dx equals to :

(A)3

π(B)

3

2π(C)

6

π(D) none

23. The function f(x) = ∫x

0t

dt satisfies [IIT - 1996]

(A) f(x + y) = f(x) + f(y) (B)

y

xf = f(x) + f(y) (C) f(xy) = f(x) + f(y) (D) none of these

24. The value ofcos2

1

x

adx

x+−∫π

π

, a > 0 is

(A) π (B) aπ (C) π/2 (D) 2π

25. The integral

−∫1 2

1 2

/

/

[ ]x nx

x+

+−

�

1

1 d

x equals:

(A) − 1/2 (B) 0 (C) 1 (D) 2 ln (1/2)

26. If Ι (m, n) = ∫ +1

0

nm )t1(t dt, then the expression of Ι(m, n) in terms of Ι(m + 1, n – 1) is

[IIT - 2003]

(A) 1m

2n

+ –

1m

n

+ Ι (m + 1, n – 1) (B)

1m

n

+ Ι (m + 1, n – 1)

(C) 1m

2n

− –

1m

n

+ Ι (m + 1, n – 1) (D)

1m

n

+ Ι (m +1, n – 1)

27. If ∫1

xsin

2t (f(t)) dt = (1 – sinx), then f

3

1 is [IIT - 2005]

(A) 1/3 (B) 1/ 3 (C) 3 (D) 3

28. ∫−

++++++0

2

23 )}1xcos()1x(3x3x3x{ dx is equal to [IIT - 2005]

(A) – 4 (B) 0 (C) 4 (D) 6

Part : (B) May have more than one options correct

29. The value of integral ∫π

0

)x(sinxf dx is

(A) π ∫π

0

)x(sinf dx (B) π ∫π 2/

0

dx)x(sinf (C) 0 (D) none of these

30. If f(x) is integrable over [1, 2], then ∫2

1

)x(f dx is equal to

(A) ∞→nlim

n

1 ∑

=

n

1rn

rf (B) ∞→n

lim n

1 ∑

+=

n2

1nrn

rf

(C) ∞→nlim

n

1 ∑

=

+n

1rn

nrf (D) ∞→n

lim n

1 ∑

=

n2

1rn

rf

31. If f(x) = ∫ +x

0

44 )tsint(cos dt, f (x + π) will be equal to

(A) f(x) + f(π) (B) f(x) + 2 f(π) (C) f(x) + f

π2

(D) f(x) + 2f

π2

32. The value of ( )2 3 3

1 2 2

2

20

1x x

x x x

+ +

+ + +∫

( ) dx is:

(A)π4

+ 2 ln2 − tan−1 2 (B)π4

+ 2 ln2 − tan−11

3(C) 2 ln2 − cot−1 3 (D) −

π4

+ ln4 + cot−1 2

33. Given f is an odd function defined everywhere, periodic with period 2 and integrable on every interval. Let

g(x) = ∫x

0

f(t) dt. Then:

(A) g(2n) = 0 for every integer n (B) g(x) is an even function(C) g(x) and f(x) have the same period (D) none

34. If Ι = ∫π

+

2/

03 xsin1

dx, then

(A) 0 < Ι < 1 (B) Ι > 2

π(C) Ι < π2 (D) Ι > 2π

35. If In = ( )

dx

xn

1 20

1

+∫ ; n ∈ N, then which of the following statements hold good?

(A) 2n In + 1

= 2 −n + (2n − 1) In

(B) I2 =

π8

1

4+

(C) I2 =

π8

1

4− (D) I

3 =

π16

5

48−

EXERCISE–8

1.

0

π

∫ e xcos2 cos3 (2n + 1) x d

x, n ∈ I

2. If f, g, h be continuous functions on [0, a] such that f (a − x) = f (x), g (a − x) = − g (x)

and 3 h (x) − 4 h (a − x) = 5, then prove that,

0

a

∫ f (x) g (x) h (x) = 0.

3. Assuming

0

π

∫ log sin x d x = − π log 2, show that,

0

π

∫ θ3 log sin θ d θ =3

2

π

0

π

∫ θ2 log ( )2 sin θ d θ.

4. Show that fa

x

x

a

x

xdx a f

a

x

x

a

dx

x( ).

lnln . ( ).

0 0

∞ ∞

∫ ∫+ = +

5. Prove that0

x

∫ f t dt

u

( )

0

∫

du =0

x

∫ f (u).(x − u) du. 6. Prove that

( )dx

x

dx

xn

nn

1 11

0

1

0 + −= ∫∫

∞

/ (n > 1)

7. Prove that Limitn → ∞

1

ncos cos cos ...... cos2 2 2 2

2

2

2

3

2 2

p p p p

n n n

π π π π+ + + +

=

p r

rr

p +

=∏

41

8.

0

π

∫( )

x dx

a x b x2 2 2 2

2cos sin+

9. Evaluate 0

1

∫ x − t. cos π t dt where ‘x’ is any real number

10.

−∫

1 3

1 3 cos tan− −

+

+

−

+

12

12

2

1

2

1

1

x

x

x

x

ex

dx

11. Evaluate, I =

0

1

∫ 2 sin (p t) sin (q

t) d

t, if:

(i) p & q are different roots of the equation, tan x = x.

(ii) p & q are equal and either is root of the equation tan x = x.

12. Prove that ∫ +

x

01x

xsin dx ≥ 0 for x ≥ 0.

13. Let f(x) be a continuous functions ∀ x ∈ R, except at x = 0 such that ∫a

0

dx)x(f , a ∈ R+ exists. If

g(x) = ∫a

xt

)t(f dt, prove that ∫

a

0

dx)x(g = ∫a

0

dx)x(f

14. If f(x) = x

xsin ∀ x ∈ (0, π], prove that,

2

π ∫

π

−π

2/

0

x2

f)x(f dx = ∫π

0

dx)x(f

15. Letd

dxF x

e

x

x

( )sin

= , x > 0. If2

2

1

4e

x

xsin

∫ dx = F (k) − F (1) then one of the possible values of k is ______.

16.−∫

1 3

1 3

x

x

4

41− cos−1 2

1 2

x

x+

dx. [IIT - 1995, 5 + 2 + 2 ]

17. Evaluate ∫π

0

|xcos|e

+

xcos

2

1cos3xcos

2

1sin2 sinx dx. [IIT - 2005, 2]

18. The value of 5050

∫

∫

−

−

1

0

10150

1

0

10050

dx)x1(

dx)x1(

is [IIT - 2006, (6, 0)]

ANSWER

EXERCISE–7

1. D 2. C 3. B 4. B 5. B 6. C

7. B 8. D 9. A 10. B 11. B 12. B

13. C 14. D 15. B 16. A 17. B 18. A

19. C 20. C 21. B 22. A 23. C 24. C

25. A 26. A 27. C 28. C 29. AB 30. BC

31. AD 32. AD 33. ABC 34. BC 35. AB

EXERCISE–8

1. 0 8. 33

222

ba4

ba )( +π

9. 2

2

π− cos πx for 0 < x < 1 ;

2

2

π for x ≥ 1 & 2

2

π− for x ≤ 0

10.π

2 311. (i) 0 (ii)

p

p

2

21 +

15. 16 16.π4

�n ( )2 312 3

2

+ + −π π

17.5

24

−

+

1

2

1sine

2

1

2

1cose 18. 5051

ASSERTION AND REASON Some questions (Assertion–Reason type) are given below. Each question contains Statement

– 1 (Assertion) and Statement – 2

(Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice :

Choices are :

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.

(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

(C) Statement – 1 is True, Statement – 2 is False.

(D) Statement – 1 is False, Statement – 2 is True.

IINNDDEEFFIINNIITTEE && DDEEFFIINNIITTEE IINNGGEEGGRRAATTIIOONN 129. Let F(x) be an indefinite integral of cos

2x.

Statement-1: The function F(x) satisfies F(x + π) = F(x) ∀ real x

Statement-2: cos2(x + π) = cos

2x.

130. Statement-1: ∫|x| dx can not be found while

1

1

| x |dx−

∫ can be found.

Statement-2: |x| is not differentiable at x = 0.

131. Statement-1: 4

1

1 x

+ ∫ dx = tan

–1 (x

2) + C Statement-2:

2

1dx

1 x+∫ = tan

–1x + C

132. Statement-1: If y is a function of x such that y(x – y)2 = x then

2dx 1log(x y) 1

x 3y 2 = − − −∫

Statement-2: dx

x 3y−∫ = log (x – 3y) + c

133. Statement–1 : f(x) = logsecx –

2x

2 Statement–2 : f(x) is periodic

134. Statement–1 :

9/ 211/ 2 11

11

x 2dx ln x 1 x c

111 x= + + +

+∫

Statement–2 : 2

2

dxln | x 1 x | c

1 x= + + +

+∫


101

0

tan x dx 10 tan1− = − ∫ ; where [x] = G.I.F.

Statement–2 : [tan–1

x] = 0 for 0 < x < tan 1 and [tan–1

x] = 1 for tan 1 ≤ x < 10.


/ 2

30

dx

41 tan x

ππ

=+

∫

Statement–2 : ( )a a

0 0

f (x) dx f a x dx= +∫ ∫

/ 2 / 2

3 30 0

dx dx

41 tan x 1 cot x

π ππ

= =+ +

∫ ∫

a a

0 0

f (x) dx f (a x) dx= −∫ ∫ .

137. Statement–1 : 2

0

1 sin xdx 0

π

− =∫ Statement–2 :

0

cos x dx 0

π

=∫ .

138. Statement–1 : ( )x 2 xe tan x sec x dx e tan x c+ = +∫

Statement–2 : ( )x xe f (x) f (x) dx e f (x) c′+ = +∫ .

139. Statement–1 : If f(x) satisfies the conditions of Rolle's theorem in [α, β], then f (x)dx

β

α

′ =β − α∫

Statement–2 : If f(x) satisfies the conditions of Rolle's theorem in [α, β], then f (x)dx 0

β

α

′ =∫


4

0

[| sin x | | cos x |]dx

π

+∫ , where [⋅] denotes G.I.F. equals 8π.

Statement–2 : If f(x) = |sinx| + |cosx|, then 1 ≤ f(x) ≤ 2 .

141. Let f(x) be a continuous function such that

n 1

3

n

f (x) dx n ,

+

=∫ n∈I

Statement–1 :

3

3

f (x)dx 27−

=∫ Statement–2 :

2

2

f (x)dx 27−

=∫

142. Let In = ( )e

n

1

nx∫ � dx, n ∈ N

Statement–I : I1. I2, I3 . . . is an increasing sequence. Statement–II : n� x is an increasing function.

143. Let f be a periodic function of period 2. Let g(x) =

x

0

f (t)∫ dt and h(x) = g(x + 2) – g(x).

Statement–1 : h is a periodic function. Statement–2 : g(x + 2) – g(x) = g(2).

144. Statement–1 : ( )x

xe1 x log x dx e log x c

x+ = +∫

Statement–2 : ( )x xe f (x) f (x) dx e f (x) c′+ = +∫ .

145. Statement–1 : If I1 = 1

2

x

dt

1 t+∫ and

1/ x

2 2

1

dtI , x 0

1 t>

+∫ then I1 = I2.

Statement–2 : { }2

2

min . x [x], x [ x] dx 0−

− − − − =∫

146. Statement–1 : 8 < 6

4

2x dx 12<∫ .

Statement–2 : If m is the smallest and M is the greatest vlaue of a function f(x) in an interval (a, b),

then the vlaue of the integral

b

a

f (x)dx∫ is such that for a < b, we have M(b – a) ≤

b

a

f (x)dx M(b a)≤ −∫ .


axax e

e sin bxdxA

=∫ (asinbx – bcosbx)+c Then A is 2 2

a b+

Statement–2 : x

2

1 sin x cos xe dx

cos x

+

∫ = ex tanx + c


2

2

d(x 1)

2

+

λ +∫ is equal to

22 x 2 c+ +

Statement–2 :

a / 2

11

xdx

1 x+∫ is 2/11 ln |x +

111 x+ | + c


/ 3

3

/ 6

1

1 tan x

π

π+∫ is π/12 Statement–2 :

b b

a a

f (x)dx f (a b x)= + −∫ ∫ dx

150. Statement–1 : If f satisfies f(x + y) = f(x) + f(y) ∀ x , y ∈R then

5

5

f (x) dx−

∫ = 0

Statement–2 : If f is an odd function then

a

a

f (x) dx−

∫ = 0

151. Statement–1 : If f(x) is an odd function of x then

x

a

f (t)dt∫ is an even function of (n)

Statement–2 : If graph of y = f(x) is symmetric about y–axis then f(x) is always an even function.

152. Statement–1 : Area bounded by y = {x}, {x} is fractional part of x = 0, x = 2 and x–axis is 1.

Statement–2 : Area bounded by y = |sinx|, x = 0, x = 2π is 2 sq. unit.

153. Statement-1: 2 2 2n

1 1 1lim ....

33n4n 1 4n 2→∞

π+ + + =

− −

Statement-2:

1n

nr 1 0

1 rlim f f (x)dx

n n→∞=

=

∑ ∫ , symbols have their usual meaning.

154. Statement-1: If In = ∫tann x dx, then 5 (I4 + I6) = tan

5x .

Statement-2: If In = ∫ tan4x dx, then

n 1tan x

n

−

- In-2 = In, n∈N.

155. Statement-1: If a > 0 and b2 – 4ac < 0, then the value of the integral

2

dx

ax bx c+ +∫ will be of the type µ tan-

1

x Ac

B

+ +

, where A, B, C, µ are constants.

Statement-2: If a > 0, b2 – 4ac < 0 then ax2 + bx + c can be written as sum of two squares.

156. Statements-1:

2 xx

2 3/ 2 2

x x 1 ee dx c

(x 1) x 1

− += +

+ +∫ Statements-2:

xe (f (x) f (x)dx′+∫ = ex f(x) + c

157. Statements-1:

2

24 2 1

x 2dx

x 2(x 5x 4) tan

x

−

−

++ +

∫ = log |tan-1

(x + 2/x)| + c

Statements-2: 1

2 2

dx 1 xtan c

a x a a

−= ++∫

158. Statements-1: 2

xln

xe c(ln x) ln x

= +∫ Statements-2: ∫ex (f(x) + f′(x)) dx = ex f(x) + c.

159. Statements-1: 43 4

1 1 1dx 1 c

2 xx 1 x= − + +

+∫ Statements-2: For integration by parts we have to follow ILATE rule.

160. Statements-1: A function F(x) is an antiderivative of a function f(x) if F ′(x) = f(x)

Statements-2: The functions x2 + 1, x

2 − π, x

2 + 2 are all antiderivatives of the function 2x.

161. Statements-1: ∫b

ax

x dx = b – a , a < b

Statements-2: If f(x) is a function continuous every where in the interval (a, b) except x = c then b c b

a a c

f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫

162. Statements-1:

3

3

1

4 3 x dx 2 30≤ + ≤∫

Statements-2: m and M be the least and the maximum value of a continuous function

y = f(x) in [a, b] then

b

a

m(b a) f (x)dx M(b a)− ≤ ≤ −∫


1

x

0

1 e dx e< <∫

Statements-2: if f(x) ≤ g(x) ≤ h(x) in (a, b) then

b b b

a a a

f (x)dx g(x)dx h(x)dx≤ ≤∫ ∫ ∫

164. Statements-1:

1

4

0

1 x dx 1.2+ <∫

Statements-2: For any functions f(x) and g(x), integrable on the interval (a,b), then

b b b

2 2

a a a

f (x)g(x)dx f (x)dx g (x)dx≤∫ ∫ ∫

165. Statements-1:

1

2

1

1dx 2

x−

= −∫

Statements-2: If F(x) is antiderivative of a continuous function (a, b) then

b

a

f (x)dx F(b) F(a)= −∫


cos x

(1 sin x)+ can be integrated by substitution it sinx = t.

Statements-2: All integrands are integrated by the method of substitution only.

167. Statement-1 : x

2

1 sin x cose dx

cos x

+

∫ = xe tan x c+∫

Statement-2 : ∫ex (f(x) + f ′ (x)dx = e

x f′(x) + c

168. Statements-1: x 2 x x x1 1

e (x 1)cos (x.e )dx x.e sin 2(x.e ) C2 4

+ = + +∫

Statements-2: ( ) { }f (x) '(x)dx, (x) tφ φ φ =∫ equals f (t)dt∫ .

169. Statements-1: log xdx x log x x c= − +∫

Statements-2: du

uvdx u vdx vdx dxdx

= +

∫ ∫ ∫ ∫

170. Statements-1:

2 xx

2 2

x 4x 2 ee dx C

x 4x 4 (x 2)

+ += +

+ + + ∫ Statements-2: ( )x xe f (x) f '(x) dx e f (x) C+ = +∫

171. Statements-1:

1 12 2

1 0

sin x x x2

3 | x | 3 | x |−

−= −

+ +∫ ∫ Statements-2:

a a a

a 0 0

f (x) dx f (x)dx f ( x)dx−

= = + −∫ ∫ ∫

172. Statements-1: The value of

1

3

0

(1 x)(1 x )dx+ +∫ can not exceed 15

8

Statements-2: If m ≤ f(x) ≤ M ∀ x ∈ [a, b] then

b

a

m(b a) f (x)dx (b a)M− ≤ ≤ −∫

173. Statements-1:

/ 2 5/ 2

5/ 2 5/ 2

0

(sin x)dx

(sin x) (cos x) 4

ππ

=+∫ Statements-2: Area bounded by y = 3x and y = x

2 is

9

2= sq. units

174. Statements-1:

9 x

e

x 10

10x 10 log 10

10 x

+

+∫ dx = log|10 x + x10

| + c Statements-2: f (x)

dx log| f (x) | cf (x)

′= +∫

175. Statements-1:

x

2 x

e (1 x)dx

cos (xe )

+∫ = tan (xe

x) + c Statements-2:

2sec xdx tan x c= +∫

176. Statement-1 : f(x) =

x

2

1

ln t dt(x 0),

1 t t>

+ +∫ then f(x) = - 1

fx

Statements-2: f(x) =

x

1

ln t dt

t 1+∫ , then f(x) + 1 1

fx 2

=

(ln x)2

177. Statement-1 :

1 12 2

1 0

sin x x 2xdx dx

3 | x | 3 | x |−

− −=

− −∫ ∫ .

Statements-2: Since sin x

3 | x |− is an odd function. So, that

1

1

sin x0

3 | x |−

=−∫

.

178. Statements-1 :

n t

0

| sin x |dx

π+

∫ = (2n + 1) – COSt (0 ≤ t ≤ π)

Statements-2:

b c b

a a c

f (x)dx f (x) dx f (x) dx= +∫ ∫ ∫ and

na a

0 0

f (x)dx n f (x) dx=∫ ∫ if f(a + x) = f(x)

179. Statements-1: The value of the integral 2

1

x

0

e dx∫ belongs to [0, 1]

Statements-2: If m & M are the lower bound and the upper bounds of f(x) over [a, b] and f is integrable, then m (b − a) ≤ b

a

f (x) dx∫ ≤ M(b – a).


0

[cot x]dx

∞

−

∫ = cot1, where [⋅] denotes greatest integer function.

Statements-2:

b

a

f (x) dx∫ is defined only if f(x) is continuous in (a, b) [⋅] function is discontinuous at all integers

181. Statements-1: ( )4

2 2

4

1 x x 1 x x dx−

+ + − − +∫ = 0 Statements-2:

a

a

f (x)dx 0−

=∫ if f(x) is an odd function.

182. Statements-1: All continuous functions are integrable

Statements-2: If a function y = f(x) is continuous on an interval [a,b] then its definite integral over [a, b] exists.

183. Statements-1: If f(x) is continuous on [a, b], a ≠ b and if

b

a

f (x)dx 0=∫ , then f(x) = 0 at least once in [a, b]

Statements-2: If f is continuous on [a, b], then at some point c in [a, b] f(c) =

b

a

1f (x)dx

b a− ∫

184. Statements-1:

4

4

| x 2 |dx 50−

+ =∫ Statements-2:

b c b

0 a c

f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫ where C∈ (A, B)

185. Statements-1:

2

2

1 xlog dx 0

1 x−

+ =

− ∫ Statements-2: If f is an odd function

a

a

f (x) dx 0−

=∫

186. Statement-1 If ax

0

1e dx

a

∞− =∫ then

m ax

m 1

0

m!x e dx

a

∞−

+=∫ Statement-2 :

nkx

n

d(e )

dx = k

n e

kx and

n n

n n 1

d 1 ( 1) n!

dx x x +

− =

187. Statement-1 :

10

0

{x [x]dx 5− =∫ Statements-2:

na a

a 0

f (x)dx n f (x)dx=∫ ∫

188. Statements-1:

0

| cos x | dx 2

π

=∫ Statements-2:

b c b

a a c

f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫ where a < c < b.

189. Statements-1:

cos x

cos x cosx

0

edx

e e

π

−= π

+∫ Statements-2:

b b

a a

f (x)dx f (a b x)dx= + −∫ ∫

190. Statements-1:

1000

x [x]

0

e dx 1000(e 1)− = −∫ Statements-2:

n 1

x [x] x [x]

0 0

e dx n e dx− −=∫ ∫

191. Statements-1: tan x

0

dx

1 2 2

ππ

=+∫

Statements-2:

b b

a a

f (x) dx f (a b x) dx= + −∫ ∫

ANSWER 129. D 130. B 131. D 132. C 133. A 134. A 135. A 136. C

137. D 138. A 139. A 140. D 141. D 142. D 143. A 144. A

145. C 146. A 147. D 148. C 149. A 150. A 151. C 152. C

153. D 154. C 155. A 156. C 157. A 158. A 159. B 160. B

161. A 162. A 163. A 164. A 165. D 166. C 167. C 168. A

169. C 170. A 171. A 172. A 173. B 174. A 175. A 176. D

177. A 178. A 179. D 180. A 181. A 182. B 183. A 184. A

185. A 186. A 187. C 188. A 189. D 190. A 191. A

Que. from Compt. Exams

(Indefinite Integral)

1. ∫ =−− )cos()cos( bxax

dx

(a) cbx

axba +

−

−−

)sin(

)sin(log)(cosec (b) c

bx

axba +

−

−−

)cos(

)cos(log)(cosec

(c) cax

bxba +

−

−−

)sin(

)sin(log)(cosec (d) c

ax

bxba +

−

−−

)cos(

)cos(log)(cosec

2. =+++

∫bxax

dx [AISSE 1989]

(a) cbxaxab

++−+−

])()[()(3

2 2/32/3 (b) cbxaxba

++−+−

])()[()(3

2 2/32/3

(c) cbxaxba

++++−

])()[()(3

2 2/32/3 (d) None of these

3. ∫ =+

+dx

xx

xx

cos5sin4

sin3cos3 [EAMCET 1991]

(a) )cos5sin4log(41

3

41

27xxx +− (b) )cos5sin4log(

41

3

41

27xxx ++

(c) )cos5sin4log(41

3

41

27xxx −− (d) None of these

4. If ∫ +−=+ acxdxxx )2sin(2

1)2cos2(sin , then the value of a and c is [Roorkee 1978]

(a) 4/π=c and ka = (an arbitrary constant) (b) 4/π−=c and 2/π=a

(c) 2/π=c and a is an arbitrary constant (d) None of these

5. ∫ =−

−−dx

x

xx

)1(

22

3

[AI CBSE 1985]

(a) cx

x

x+−

−

+

21

1log

2

(b) cx

x

x++

+

−

21

1log

2

(c) cx

x

x++

−

+

21

1log

2

(d) cx

x

x+−

+

−

21

1log

2

6. ∫ =−

−dx

xx

xx22

88

cossin21

cossin [IIT 1986]

(a) cx +2sin (b) cx +− 2sin2

1 (c) cx +2sin

2

1 (d) cx +− 2sin

7. =+∫ 2

2

)( bxa

dxx [IIT 1979]

(a)

+−++

bxab

abxa

b

ax

b

1)log(

212

2 (b)

+++−

bxab

abxa

b

ax

b

1)log(

212

2

(c)

++++

bxab

abxa

b

ax

b

1)log(

21 2

2 (d)

+−+−+

bxab

abxa

b

a

b

ax

b

1)log(

21 2

2

8. =++

∫ − 21222 )(tan)1( xqpx

dx

(a) cxqpxqq

+++ −− ])(tantanlog[1 21221 (b) cxqpxq +++ −− ])(tantanlog[ 21221

(c) cxqpq

++ − 2/3122 )tan(3

2 (d) None of these

9. ∫ =+

dx

x

x

3

5

1

[IIT 1985]

(a) cx ++ 2/33 )1(9

2 (b) cxx ++++ 1 2/32/33 )1(

3

2)1(

9

2

(c) cxx ++−+ 2/132/33 )1(3

2)1(

9

2 (d) None of these

10. ∫+− 2cossin xx

dx equals [MP PET 2002]

(a) cx

+

+−

82tan

2

1 π (b) c

x+

+

82tan

2

1 π (c) c

x+

+

82cot

2

1 π (d) c

x+

+−

82cot

2

1 π

11. =+∫ x

ceb

dxa [MP PET 1988; BIT Ranchi 1979]

(a) cceb

e

b

ax

x

+

+log (b) c

e

ceb

b

ax

x

+

+log (c) c

ceb

e

a

bx

x

+

+log (d) c

e

ceb

a

bx

x

+

+log

12. =∫ dxxsin [Roorkee 1977]

(a) cxx +− ]cos[sin2 (b) cxxx +− ]cos[sin2

(c) cxx ++ ]cos[sin2 (d) cxxx ++ ]cos[sin2

13. ∫ =−

dxx

x2/32

2

)9(

(a) cx

x

x+−

−

−

3sin

9

1

2 (b) c

x

x

x++

−

−

3sin

9

1

2 (c) c

x

xx+

−−−

2

1

93sin (d) None of these

14. =+

−∫ dx

x

xx

2

2

1

1

(a) cxx +−+− ]1[sin2

1 421 (b) cxx +−+− ]1[sin2

1 221

(c) cxx +−+− 421 1sin (d) cxx +−+− 221 1sin

15. If cxfab

dxxxxf +−

=∫ ))(log()(2

1cossin)(

22, then =)(xf

(a) xbxa

2222 cossin

1

+ (b)

xbxa2222 cossin

1

− (c)

xbxa2222 sincos

1

+ (d)

xbxa2222 sincos

1

−

16. ∫ =+ xx

dx22 cos5sin4

[AISSE 1986]

(a) cx

+

−

5

tan2tan

5

1 1 (b) cx

+

−

5

tantan

5

1 1 (c) cx

+

−

5

tan2tan

52

1 1 (d) None of these

17. ∫ =+−

+dx

xx

x

1

124

2

[MP PET 1991]

(a) cx

x+

+−2

1 1tan (b) c

x

x+

+−2

1 1cot (c) c

x

x+

−− 1tan

21 (d) c

x

x+

−− 1cot

21

18. ∫ =dxx 2)(log [IIT 1971, 77]

(a) cxxxxx +−− 2log2)(log 2 (b) cxxxxx +−− log2)(log 2

(c) cxxxxx ++− 2log2)(log 2 (d) cxxxxx ++− log2)(log 2

19. The value of ∫−

dxx

ax )( 22

will be [UPSEAT 1999]

(a)

−−− −

a

axaax

)(tan)(

22122 (b)

−+− −

a

axaax

)(tan)(

22122

(c) ][tan)( 221222 axaax −+− − (d) cax +− /tan 1

20. =∫ dxxx 2sec2tan 3 [IIT 1977]

(a) cxx +− 2sec2

12sec

6

1 3 (b) cxx ++ 2sec2

12sec

6

1 3

(c) cxx +− 2sec3

12sec

9


21. =∫− dxxx 1sin [MP PET 1991]

(a) cxx

xx

+−+

− − 21

2

14

sin4

1

2 (b) cx

xx

x+−+

+ − 21

2

14

sin4

1

2

(c) cxx

xx

+−−

− − 21

2

14

sin4

1

2 (d) cx

xx

x+−−

+ − 21

2

14

sin4

1

2

22. ∫ =−

dxx

xa

(a) ca

xa

a

x

a

xa +

−+−1sin (b) cxa

a

x

a

x+−+− 221sin

(c) cxaa

x

a

xa +

−−− 221sin (d) cxa

a

x

a

x+−−− 221sin

23. If

∈

4

3,

4

ππx , then ∫ =

−

−dxxe

x

xx x cos2sin1

cossin sin

(a) ce x +sin (b) ce xx +−cossin

(c) ce xx ++cossin (d) ce xx +−sincos

24. If CeBAxdxee

ee x

xx

xx

+−+=−

+∫ −

−

)49log(49

64 2 , then A, B and C are [IIT 1990]

(a) +=== 3log2

3,

35

36,

2

3CBA constant

(b) +=== 3log2

3,

36

35,

2

3CBA constant

(c) +−=−=−= 3log2

3,

36

35,

2

3CBA constant

(d) None of these

25. The value of ∫ dxx3sec will be [UPSEAT 1999]

(a) [ ])tanlog(sectansec2

1xxxx ++

(b) [ ])tanlog(sectansec3

1xxxx ++

(c) [ ])tanlog(sectansec4

1xxxx ++

(d) [ ])tanlog(sectansec8

1xxxx ++

26. ∫ =+

−dxe

x

x x

3)1(

1 [IIT 1983; MP PET 1990]

(a) cx

ex

++

−2)1(

(b) cx

ex

++ 2)1(

(c) cx

ex

++ 3)1(

(d) cx

ex

++

−3)1(

27. If ∫= dxxeIx 2sin , then for what value of K, +−= )2cos22(sin xxeKI x constant [MP PET 1992]

(a) 1 (b) 3 (c) 5 (d) 7

28. The value of ∫ −− 223 xx

dx will be [UPSEAT 1999]

(a)

−

+

x

x

1

3log

4

1 (b)

−

+

x

x

1

3log

3

1

(c)

−

+

x

x

1

3log

2

1 (d)

+

−

x

x

3

1log

29. =+∫ dxxx 32 [AISSE 1985]

(a) cxxx

++−+ 2/52/3 )32(15

1)32(

3

(b) cxxx

++++ 2/52/3 )32(15

1)32(

3

(c) cxxx

++++ 2/52/3 )32(6

1)32(

2

(d) None of these

30. ∫ =

−

+θ

θθ

θθθ d

sincos

sincoslog2cos [IIT 1994]

(a)

−

+−

θθ

θθθθ

sincos

sincoslog)sin(cos 2

(b)

−

++

θθ

θθθθ

sincos

sincoslog)sin(cos 2

(c)

+

−−

θθ

θθθθ

sincos

sincoslog

2

)sin(cos 2

(d) θθπ

θ 2seclog2

1

4tanlog2sin

2

1−

+

31. ∫ =+

dxxxx

x2

2

)cossin( [MNR 1989; RPET 2000]

(a) xxx

xx

cossin

cossin

+

+ (b)

xxx

xxx

cossin

cossin

+

−

(c) xxx

xxx

cossin

cossin

+

− (d) None of these

32. If ∫= dxbxeuax cos and ∫= dxbxev

ax sin , then =++ ))(( 2222vuba

(a) axe2 (b) axeba

222)( +

(c) axe

2 (d) axeba

222 )( −

33. If ,)(log∫= dxxIn

n then =+ −1nn nII

[Karnataka CET 2003]

(a) nxx )(log (b) n

xx )log(

(c) 1)(log −nx (d) n

xn )(log

34. ∫ =

+ dx

xe

x

42sin2/ π

[Roorkee 1982]

(a) cx

e x +2

cos2/ (b) cx

e x +2

cos2 2/

(c) cx

ex +

2sin2/ (d) c

xe

x +2

sin2 2/

35. If Axxdxxx

x+−−−=

+−

+∫ )2ln(7)3ln(9

65

322

, then =A [MP PET 1992]

(a) +− )2ln(5 x constant (b) +−− )3ln(4 x constant

(c) Constant (d) None of these

36. ∫ =+ x

dx

cos2

(a) cx

+

−

2tan

3

1tan2 1 (b) c

x+

−

2tan

3

1tan

3

2 1

(c) cx

+

−

2tan

3

1tan

3


37. ∫ ++dx

xx

x

124 equal to [MP PET 2004]

(a)

+−

3

12tan

3

1 21 x

(b)

+−

3

12tan

3

1 21 x

(c) )12(tan3

1 21 +−x (d) None of these

38. ∫ =+ )2sin(sin xx

dx [IIT 1984]

(a) )cos21log(3

2)cos1log(

2

1)cos1log(

6

1xxx +−++−

(b) )cos21log(3

2)cos1log(2)cos1log(6 xxx +−++−

(c) )cos21log(3

2)cos1log(

2

1)cos1log(6 xxx ++++−

(d) None of these

39. If Axxxdxxx

x ae +−

+−=+−

+ −

∫122

5

2tan

2

1)1()1(log

)1)(1(

32,

where A is any arbitrary constant, then the value of ‘a’ is [MP PET 1998]

(a) 5/4 (b) – 5/3

(c) – 5/6 (d) – 5/4

40. If ,2

2

1

1log

)1()4(

)12(22

2

Cx

x

x

x

xx

dxxba

+

+

−

−

+=

−−

+∫ then the values of a and b are respectively [Roorkee 2000]

(a) 1/2, 3/4 (b) –1, 3/2

(c) 1, 3/2 (d) –1/2, ¾

(Definite Integral)

1. If I is the greatest of the definite integrals

,cos1

0

21 ∫

−= dxxeI x dxxeI x 21

02 cos

2

∫−=

,1

03

2

∫−= dxeI

x ,1

0

2/4

2

∫−= dxeI

x then

(a) 1II = (b) 2II =

(c) 3II = (d) 4II =

2. Let )(xf be a function satisfying )()( xfxf =′ with 1)0( =f and )(xg be the function satisfying .)()( 2xxgxf =+ The value of

integral ∫1

0)()( dxxgxf is equal to

[AIEEE 2003; DCE 2005]

(a) )7(4

1−e (b) )2(

4

1−e

(c) )3(2

1−e (d) None of these

3. If ∫=x

mm dxxI

1)(log satisfies the relation ,1−−= mm lIkI then

(a) ek = (b) ml =

(c) e

k1

= (d) None of these

4. Let f be a positive function. Let

{ }dxxxfxIk

k∫ −−=

11 )1( , { }dxxxfI

k

k∫ −−=

12 )1(

when .012 >−k Then 21 / II is [IIT 1997 Cancelled]

(a) 2 (b) k

(c) 2/1 (d) 1

5. If ∫∫ +=1

0,)()(

x

x

dttftxdttf then the value of )1(f is

[IIT 1998; AMU 2005]

(a) 1/2 (b) 0

(c) 1 (d) –1/2

6. ∫−

1

0 4

7

1

dx

x

x is equal to [AMU 2000]

(a) 1 (b) 3

1

(c) 3

2 (d)

3

π

7. If n is any integer, then ∫ =+π

0

3cos)12(cos

2

dxxnex

[IIT 1985; RPET 1995; UPSEAT 2001]

(a) x (b) 1 (c) 0 (d) None of these

8. The value of the definite integral ∫ +

1

03 16x

dxx lies in the interval ].,[ ba The smallest such interval is

(a)

17

1,0 (b) [0, 1]

(c)

27

1,0 (d) None of these

9. Let a,b,c be non-zero real numbers such that ∫∫ +++=+++2

0

281

0

28 ))(cos1())(cos1( dxcbxaxxdxcbxaxx

Then the quadratic equation 02 =++ cbxax has

[IIT 1981; CEE 1993]

(a) No root in (0, 2)

(b) At least one root in (0, 2)

(c) A double root in (0, 2)

(d) None of these

10. If ∫−=x

dttxf1

,||)( ,1−≥x then [MNR 1994]

(a) f and f ′ are continous for 01 >+x

(b) f is continous but f ′ is not continous for 01 >+x

(c) f and f ′ are not continous at 0=x

(d) f is continous at 0=x but f ′ is not so

11. Let ∫=x

dttfxg0

)()( where ]1,0[,1)(2

1∈≤≤ ttf and

2

1)(0 ≤≤ tf for ]2,1(∈t , then [IIT Screening 2000]

(a) 2

1)2(

2

3<≤− g (b) 2)2(0 <≤ g

(c) 2

5)2(

2

3≤< g (d) 4)2(2 << g

12. The value of ∫−>

+

π

π,0,

1

cos 2

adxa

xx

is

[IIT Screening 2001; AIEEE 2005]

(a) π (b) πa

(c) 2

π (d) π2

13. If ∫ −−=

+=

)(

)(1 )}1({,

1)(

af

afx

x

dxxxxgIe

exf , and ∫ −

−=)(

)(2 ))}1({

af

afdxxxgI , then the value of

1

2

I

I is

[AIEEE 2004]

(a) 1 (b) –3

(c) –1 (d) 2

14. Let RRf →: and RRg →: be continuous functions, then the value of the integral

∫− =−−−+2/

2/)]()([)]()([

π

πdxxgxgxfxf

[IIT 1990; DCE 2000; MP PET 2001]

(a) π (b) 1

(c) 1− (d) 0

15. The numbers P, Q and R for which the function RxQePexfxx ++= 2)( satisfies the conditions ,1)0( −=f 31)2(log =′f and

∫ =−4log

0 2

39])([ dxRxxf are given by

(a) ,2=P ,3−=Q 4=R (b) ,5−=P ,2=Q 3=R

(c) ,5=P ,2−=Q 3=R (d) ,5=P ,6−=Q 3=R

16.

+

∑∫∑∫

=

+

=−−

10

1

12

2

2710

1

2

12

27sinsin

n

n

nn

n

ndxxdxx equals

[MP PET 2002]

(a) 227 (b) 54−

(c) 36 (d) 0

17. Let ∫ =1

0,1)( dxxf ∫ =

1

0)( adxxfx and ∫ =

1

0

22 ,)( adxxfx then the value of ∫ =−1

0

2 )()( dxxfax [IIT 1990]

(a) 0 (b) 2a

(c) 12 −a (d) 222 +− aa

18. Given that ,))()((2))()((0 222222

2

∫∞

+++=

+++ accbbacxbxax

dxx π then the value of ∫∞

++0 22

2

)9)(4( xx

dxx is

[Karnataka CET 1993]

(a) 60

π (b)

20

π

(c) 40

π (d)

80

π

19. If ∫ +=1

0,)1(),( dtttnml nm then the expression for ),( nml in terms of )1,1( −+ nml is [IIT Screening 2003]

(a) )1,1(11

2−+

+−

+nml

m

n

m

n

(b) )1,1(1

−++

nmlm

n

(c) )1,1(11

2−+

++

+nml

m

n

m

n

(d) )1,1(1

−++

nmln

m

20. 5

444....321

limn

n

n

++++

∞→=

++++−

∞→ 5

333....321

limn

n

n

[AIEEE 2003]

(a) 30

1 (b) Zero

(c) 4

1 (d)

5

1

21. If ,0,5

2)(

5

0

2

>=∫ ttdxxxft

then =

25

4f

[IIT Screening 2004]

(a) 5

2 (b)

2

5

(c) 5

2− (d) None of these

22. For which of the following values of m, the area of the region bounded by the curve 2xxy −= and the line mxy = equals

2

9

[IIT 1999]

(a) 4− (b) 2−

(c) 2 (d) 4

23. Area enclosed between the curve 32 )2( xxay =− and line ax 2= above x-axis is [MP PET 2001]

(a) 2aπ (b)

2

32

aπ

(c) 22 aπ (d) 2

3 aπ

24. What is the area bounded by the curves 922 =+ yx and xy 8

2 = is [DCE 1999]

(a) 0 (b)

−+ −

3

1sin9

2

9

3

22 1π

(c) π16 (d) None of these

25. The area bounded by the curves 1|| −= xy and 1|| +−= xy is [IIT Screening 2002]

(a) 1 (b) 2

(c) 22 (d) 4

26. The volume of spherical cap of height h cut off from a sphere of radius a is equal to [UPSEAT 2004]

(a) )3(3

2 hah −π

(b) )2)(( 22ahhaha −−−π

(c) 3

3

4h

π (d) None of these

27. If for a real number ][, yy is the greatest integer less than or equal to ,y then the value of the integral ∫2/3

2/

]sin2[

π

π

dxx is

[IIT 1999]

(a) π− (b) 0

(c) 2

π− (d)

2

π

28. If ,2

sin)( Bx

Axf +

=

π 2

2

1=

′f and ∫ =

1

0,

2)(

π

Adxxf then the constants A and B are respectively [IIT 1995]

(a) 2

π and

2

π (b)

π

2 and

π

3

(c) π

4 and 0 (d) 0 and

π

4−

29. If ∫∞

−−=0

1 ,dxxeInx

n then ∫∞

−− =0

1dxxe

nxλ

(a) nIλ (b) nIλ

1

(c) n

nI

λ (d) n

nIλ

30. ∫=4/

0tan

π

dxxI nn , then ][lim 2−

∞−+ nn

nIIn equals

[AIEEE 2002]

(a) 1/2 (b) 1

(c) ∞ (d) 0

31. The area bounded by the curves xy ln= , ||ln xy = , |ln| xy = and |||ln| xy = is [AIEEE 2002]

(a) 4 sq. unit (b) 6 sq. unit

(c) 10 sq. unit (d) None of these

32. dxx

xn

∫

+π

0sin

2

1sin

, )( Nn ∈ equals [Kurukshetra CEE 1998]

(a) πn (b) 2

)12(π

+n

(c) π (d) 0

33. If ∫ =−1

0,0)(

2

dxxex α then

[MNR 1994; Pb. CET 2001; UPSEAT 2000]

(a) 21 << α (b) 0<α

(c) 10 << α (d) None of these

34. ∫π

π

10

|sin| dxx is [AIEEE 2002]

(a) 20 (b) 8

(c) 10 (d) 18

35. ∫− +

+π

πdx

x

xx2cos1

)sin1(2 is [AIEEE 2002]

(a) 4/2π (b) 2π

(c) 0 (d) 2/π

36. On the interval ,4

7,

3

5

ππ the greatest value of the function ∫ =−=

x

dtttxf3/5

)sin2cos6()(π

(a) 12233 ++ (b) 12233 −−

(c) Does not exist (d) None of these

37. If ∫∫∫ ===2

13

1

02

1

01

232

2,2,2 dxIdxIdxI xxx , ∫=2

14

3

2 dxI x , then [AIEEE 2005]

(a) 43 II = (b) 43 II >

(c) 12 II > (d) 21 II >

38. If xx

fxf =

−

13)(2 , then dxxf∫

2

1)( is equal to

[J & K 2005]

(a) 2ln5

3 (b) )2ln1(

5

3+

−

(c) 2ln5

3− (d) None of these

89 of 89

39. If 03 =∫b

adxx and

3

22 =∫ dxxb

a, then the value of a and b will be respectively [AMU 2005]

(a) 1, 1 (b) 1,1 −−

(c) 1,1 − (d) 1,1−

40. The sine and cosine curves intersects infinitely many times giving bounded regions of equal areas. The area of one of such

region is [DCE 2005]

(a) 2 (b) 22

(c) 23 (d) 24

(Indefinite Integral)

1 b 2 b 3 a 4 a 5 d

6 b 7 d 8 a 9 c 10 d

11 a 12 b 13 a 14 a 15 a

16 c 17 c 18 c 19 a 20 a

21 a 22 a 23 a 24 d 25 a

26 b 27 c 28 a 29 a 30 d

31 c 32 c 33 a 34 d 35 c

36 b 37 b 38 a 39 d 40 a

(Definite Integral)

1 d 2 d 3 b 4 c 5 a

6 b 7 c 8 a 9 b 10 a

11 b 12 c 13 d 14 d 15 d

16 d 17 a 18 a 19 a 20 d

21 a 22 b 23 b 24 b 25 b

26 a 27 c 28 c 29 c 30 b

31 a 32 c 33 c 34 d 35 b

36 b 37 d 38 b 39 d 40 b

Documents

Maths Study Material - Elementary Definite Integral