1
Indefinite IntegralsIndefinite IntegralsIndefinite IntegralsIndefinite Integrals
Indefinite Integrals Definition Indefinite Integrals Definition Indefinite Integrals Definition Indefinite Integrals Definition
If d
dx[ g (x) ] = f (x), then g (x) + c is an indefinite integral of f (x) and
we write it as ( )f x dx∫ = g (x) + c.
e.g.
2
2
d( x ) 2x
dx
2x dx x c
=
∴ = +∫
ImportanImportanImportanImportant t t t
The results which are true for x are true for ax + b also ( as both are
linear ), the only change is, obtained answer is to be divided by a ( i.e.
coefficient of x )
For example For example For example For example
2sec x dx tan x c= +∫
∴ 2 tan (x / 2)x x
sec dx c 2 tan c2 1/ 2 2
= + = +
∫
2
Different typesDifferent typesDifferent typesDifferent types
Type Type Type Type –––– I I I I
Sums based on standard result.
Type Type Type Type –––– IIIIIIII
Integrand of the type P (x)
ax b+ , where P (x) is a polynomial
Here divide numerator by denominator and use
log |ax b |dxc
ax b a
+= +
+∫
Type Type Type Type ---- IIIIIIIIIIII
Integration by substitutionIntegration by substitutionIntegration by substitutionIntegration by substitution
'( )log ( )
( )
f xdx f x c
f x= +∫
Standard SubstitutionsStandard SubstitutionsStandard SubstitutionsStandard Substitutions
Expression Substitution 2 2a x− x = a sin θ or x = a cos θ 2 2
a x+ x = a tan θ or x = a cot θ 2 2
x a− x = a sec θ or x = a cosec θ
a x
a x
−
+ x = a cos θ
3
Type Type Type Type –––– IVIVIVIV
Integrals of the type
x
x
a sin x b cos x a e bdx and dx
csin x d cos x ce d
+ +
+ +∫ ∫
Here we write
dNumerator A ( Denominator ) B ( Denominator)
dx= +
Values of A and B are to be obtained by equating coefficients of sin x
and cos x.
Type Type Type Type –––– VVVV
To express angle in numerator in terms of angle in the denominator.
e.g.
To evaluate sin ( x a )
dxsin ( x a )
−
+∫
We write
x – a = ( x + a ) – 2a
Type Type Type Type –––– VIVIVIVI
Integral of type
2 22
px q px q P(x )dx OR dx OR dx
ax bx c ax bx cax bx c
+ +
+ + + ++ +∫ ∫ ∫
Here we write
2dpx q A B ( ax bx c )
dx+ = + + +
In the third integral P(x) is a polynomial of degree greater than or
equal to 2. Here, divide numerator by denominator and then proceed.
4
Note Note Note Note : In CET, values of A and B can be obtained directly.
e.g. in 2
4x 3dx
x 2x 1
+
+ +∫
As derivative of x 2 + 2x + 1 is 2x + 2
we write 4x + 3 = 2 ( 2x + 2 ) – 4 + 3 = 2 ( 2x + 2 ) – 1
Type Type Type Type –––– VIIVIIVIIVII ( Important )( Important )( Important )( Important )
Integral of the type 1 1 1
dx OR dx OR dxa b sin x a b cos x a b sin x ccos x+ + + +∫ ∫ ∫
Here we put x
tan t2
=
2
2 2 2
2dt 1 t2tdx , sin x , cos x
1 t 1 t 1 t
−∴ = = =
+ + +
Note :Note :Note :Note :
If angle is 2x, put tan x = t 2
2 2 2
1 tdt 2tdx , sin x , cos x
1 t 1 t 1 t
−∴ = = =
+ + +
5
Type Type Type Type –––– VIIVIIVIIVIIIIII
Integral of the type
2 2 2 2
1 1 1dx OR dx OR dx
a b sin x a b cos x a b sin x c cos x+ + + +∫ ∫ ∫
Here we multiply numerator and denominator by sec 2 x and put
tan x = t
Note :Note :Note :Note :
In the denominator
sin 2 x × sec 2 x = tan 2 x
cos 2 x × sec 2 x = 1
sec 2 x = 1 + tan 2 x
Type Type Type Type –––– IXIXIXIX
Integration By partsIntegration By partsIntegration By partsIntegration By parts
The Theorem is
duuv dx u v dx v dx dx
dx
= −
∫ ∫ ∫ ∫
This can be remembered as
fis – idfis Rule
The order in which u and v are to be taken is according to the serial
order of the letters of the word “LIATE”, where
L : Logarithmic
I : Inverse Trigonometric
A : Algebraic
T : Trigonometric
E : Exponential
6
Type Type Type Type –––– XXXX
Sums on
1. 2 2sin ( ) ( sin cos )
a xa x e
e bx dx a bx b bx ca b
= − ++∫
2. 2 2cos ( ) ( cos sin )
a x
a x ee bx dx a bx b bx c
a b= + +
+∫
Type Type Type Type –––– XIXIXIXI
Sums based on x xe [f (x) f '(x)] dx e f (x) c+ = +∫
Here multiple of e x is expressed as sum of a function and its
derivative.
Typical sums of this type 2
1tan x
2 2
log x 1 x xdx , e dx
(1 log x ) 1 x
− + +
+ +∫ ∫
7
Type Type Type Type –––– XIIXIIXIIXII
Partial FractionsPartial FractionsPartial FractionsPartial Fractions
1. 1. 1. 1. Distinct Linear FactorsDistinct Linear FactorsDistinct Linear FactorsDistinct Linear Factors
ExampleExampleExampleExample
2
3x 5 3x 5 A B
( x 1) ( x 3) ( x 1) ( x 3)x 2x 3
+ += = +
− + − ++ −
We write
3x + 5 = A ( x + 3 ) + B ( x – 1 )
And so on
Disguised linear factorsDisguised linear factorsDisguised linear factorsDisguised linear factors
Example Example Example Example
In
2
2 2
x
( x 4) ( x 7 )+ − we take x 2 = t for finding partial fractions
only. This is not a substitution.
2. 2. 2. 2. Repeated Linear FactorsRepeated Linear FactorsRepeated Linear FactorsRepeated Linear Factors
ExampleExampleExampleExample
2 2
x 1 A B C
( x 1) ( x 2)( x 1) ( x 2 ) ( x 1)
+= + +
− +− + −
8
3.3.3.3. Non Non Non Non –––– repeated repeated repeated repeated quadraticquadraticquadraticquadratic factorfactorfactorfactor
Example Example Example Example
2 2
Bx cx A
( 2x 1)( 2x 1) ( x 3) ( x 3)
+= +
++ + +
Type Type Type Type –––– XIIIXIIIXIIIXIII
Reduction FormulaeReduction FormulaeReduction FormulaeReduction Formulae
1. 1 21 1
sin sin cos sinn n nnx dx x x x dx
n n
− −− −= +∫ ∫
2. 1 21 1
cos cos sin cosn n nnx dx x x x dx
n n
− −−= +∫ ∫
Poll QuestionPoll QuestionPoll QuestionPoll Question
log f (x)e f (x)=
9
Proof :Proof :Proof :Proof :
y
e
y
log f (x)e
Let e f (x)
y log f (x)
Put this in e f (x) to get
e f (x)
=
∴ =
=
=
Solved SumsSolved SumsSolved SumsSolved Sums
CET CET CET CET –––– 2008 ( Memory Based )2008 ( Memory Based )2008 ( Memory Based )2008 ( Memory Based )
1. 2 2
3 dx
( x 1) ( x 4)=
+ +∫
(a) log ( x 2 + 1 ) – log ( x 2 + 4 ) + c
(b) 1 11 x
tan x tan c2 2
− − − +
(c) 1 1 x
2 tan x tan c2
− − − +
(d) 1 11
tan x tan x c2
− −− +
SolutionSolutionSolutionSolution
2 2 2 2
1 1
3 dx 1 1I dx
( x 1) ( x 4) x 1 x 4
1 xtan x tan c
2 2
− −
= = −
+ + + +
= − +
∫ ∫
2. 2
1 1dx
log x ( log x )
− =
∫
10
(a) 1
clog x
+ (b) x
clog x
+ (c) ( ) 2
xc
log x+ (d) log x + c
Solution Solution Solution Solution
2
1 1I dx
log x ( log x )
= −
∫
Put log x = t ∴ x = e t ∴ dx = e t dt
t t
2
1 1 1 xI e dt e c c
t t log xt
∴ = − = + = +
∫
3.
x
3
x edx
( x 2)=
+∫
(a)
xec
x 2+
+ (b)
( )
x
2
ec
x 2+
+
(c) ( )
x
3
ec
x 2+
+ (d)
( )
x
2
x ec
x 2+
+
Solution Solution Solution Solution
( )
x xx
3 3 2 3
x
2
x e e ( x 2 2) 1 2I dx dx e dx
( x 2) ( x 2) ( x 2) ( x 2)
ec
x 2
+ −= = = −
+ + + +
= ++
∫ ∫ ∫
4. 2 2 2
1dx
a b x=
−∫
(a) 11 bx
sin cb a
− +
(b)
11 axsin c
ab b
− +
(c) 11 ax
sin cb b
− +
(d)
11 bxsin c
ab a
− +
11
SolutionSolutionSolutionSolution
2 2 2 2
2
1
1 1 1I dx dx
ba b x ax
b
1 bxsin c
b a
−
= =−
−
= +
∫ ∫
5. n
1dx
x x −=
+∫
(a) n 11
log | x 1| cn 1
++ +
+ (b)
n1log | x | c
n+
(c) n1
log | x | cn 1
++
(d) n 11
log | x | cn
++
SolutionSolutionSolutionSolution
n
n n 1
n
n
n 1
n 1
x1 1I dx dx dx
1x x x 1x
x
( n 1 ) x1dx
n 1 x 1
1log | x 1 | c
n 1
− +
+
+
= = =+ +
+
+=
+ +
= + ++
∫ ∫ ∫
∫
CET CET CET CET –––– 2009 ( Memory Based)2009 ( Memory Based)2009 ( Memory Based)2009 ( Memory Based)
1. x
2
x 1e dx
x
−=
∫
(a)
x
2
ec
x+ (b)
xec
x+ (c)
xx e c+ (d) x 2e x c+
12
SolutionSolutionSolutionSolution
x x
2
1 1 1I e dx e c
x xx
= − = × +
∫
2. x log x dx =∫
(a) ( )x
2 log x 1 c4
− + (b) ( )2x
2 log x 1 c4
− +
(c) ( )2x
2 log x 1 c4
+ + (d) ( )x
2 log x 1 c4
+ +
SolutionSolutionSolutionSolution
( )
2 2
2 2 2
I x log x dx log x . x dx
x x1log x
2 x 2
x x x1log x 2 log x 1 c
2 2 2 4
= =
= × − ×
= − × = − +
∫ ∫
∫
CET CET CET CET –––– 2010 ( Memory Based )2010 ( Memory Based )2010 ( Memory Based )2010 ( Memory Based )
1. 2
dx
16 x 9=
+∫
(a) 11 4x
tan c4 3
− +
(b)
11 4xtan c
12 3
− +
(c) 11 4x
tan c3 3
− +
(d)
1 4xtan c
3
− +
SolSolSolSolutionutionutionution
13
22
1
1
dx 1 dxI
91616 x 9x
16
11 xtan c
316 3/ 4
4
1 4xtan c
12 3
−
−
= =+
+
= × +
= +
∫ ∫
2. tan x 2 3e (sec x sec x sin x ) dx+ =∫
(a) e x sec x tan x + c (b) e tan x . tan x + c
(c) e tan x sec 2 x + c (d) e tan x . tan 2 x + c
Solution Solution Solution Solution
tan x 2 3 tan x 2
tan x 2
2
x t tan x
e (sec x sec x sin x ) dx e sec x ( 1 sec x sin x ) dx
e ( 1 tan x)sec x dx
put tan x t sec x dx dt
I e ( t 1) dt e . t e . tan x c
+ = +
= +
= ∴ =
∴ = + = = +
∫ ∫
∫
∫
More Solved SumsMore Solved SumsMore Solved SumsMore Solved Sums
1.
5 log 4 log
3 log 2 log
x x
x x
e edx
e e
−
−∫
(a) log | x 3 – x 2 | + c (b)
3
3
xc+
(c) 2
xc+ (d) log | x ( x – 1 ) | + c
14
SolutionSolutionSolutionSolution
( ) ( )
( )
( )
x
45 4
3 2 2
32
5 4
3 2
log x log x
log x log x
log f
e eI
e e
using e f x
x x 1x xI
x x x x 1
xx dx
3
-=
-
=
--= =
- -
= =
ò
ò ò
ò
2. cos 2 cos 2
cos cos
xdx
x
α
α
−
−∫ is
(a) sin x – x sin α + c (b) x cos α + cos x + c
(c) 2 ( sin x + x cos α ) + c (d) cos x + x sin α + c
SolutionSolutionSolutionSolution
( )
( )
( ) ( )
( ) ( ) ( )
2 2
2 2
2cos x 1 2cos 1I dx
cos x cos
2 cos x cosdx
cos x cos
cos x cos cos x cos2
cos x cos
2 cos x cos dx 2 sin x x cos c 2 sin x x cos c
− − α −=
− α
− α=
− α
+ α − α=
− α
= + α = + α + = + α +
∫
∫
∫
∫
3. 1 2 tan (sec tan )x x x dx+ +∫ is
(a) 3/23
[1 2 tan (sec tan ) ]2
x x x c+ + +
(b) log Gsec x ( sec x + tan x ) G + c
(c) log Gsec x G - log G sec x + tan x G + c
(d) log G sec x + tan x G + c
15
SolutionSolutionSolutionSolution
( )
( ) ( )
[ ] [ ]
( )
2
2 2
2 2
2
I 1 2 tan x sec x 2 tan x
1 tan x 2sec x tan x tan x
sec x 2sec x tan x 2 tan x dx
sec x tan x dx sec x tan x dx
log sec x tan x log sec x c
log sec x sec x tan x 1 c
= + +
= + + +
= + +
= + = +
= + + +
= + +
∫
∫
∫
∫ ∫
4. 4 4
sin 2
cos sin
x dx
x x+∫ =
(a)
4
4
coslog
sin
x
x + c (b) cot – 1 ( tan x ) + c
(c) tan – 1 ( tan 2 x ) + c (d) 2 tan – 1 ( tan x ) + c
SolutionSolutionSolutionSolution
( )
( ) ( )
4 4
4
2 2
4 4
2
22
2 2
1 1 2
2
2sin x cos xI dx
cos x sin x
Divide Numerator and Deno min ator by cos x
sin x 12 dx
cos x cos x 2 tan x sec x dx
1 tan x tan x 1
2 tan x sec x dxI
tan x 1
put tan x t 2 tan x sec x dx dt
dtI tan t c tan tan x c
t 1
− −
=+
= =+ +
=
+
= ∴ =
= = + = ++
∫
∫ ∫
∫
∫
5. cos sin
1 sin 2
x xdx
x
−
+∫
16
(a) 1
cos sinc
x x
−+
+ (b)
1
cos sinc
x x+
−
(c) 1
sin 2c
x+ (d)
1
cos 2c
x+
SolutionSolutionSolutionSolution
( )
( )
( )2 2 2
2
2
2
1
put cos x sin x t ........ 1
sin x cos x dx dt
on squaring 1
cos x sin x 2 sin x cos x t
1 sin 2 x t
dtI t dt
t
t 1I c
1 cos x sin x
−
−
+ =
∴ − + =
+ + =
∴ + =
= =
−= = +
− +
∫ ∫
6. log( 2) log
( 2)
x xdx
x x
+ −
+∫
(a)
2
1 2log
4
xc
x
++
(b) [log(x + 2) – log x ] 2 + c
(c)
2
1 2log
4
xc
x
− ++
(d) [log(x + 2) – log x ] + c
SolutionSolutionSolutionSolution
17
( )
( )
( )2
2
2
put log x 2 log x dt
1 1d x dt
x 2 x
x x 2dx dt
x x 2
dx 1dt
x x 2 2
t1 1I t dt
2 2 2
1log x 2 log x c
4
x 21log c
4 x
+ − =
∴ − =
+
− −∴ =
+
−∴ =
+
− −∴ = × = ×
−= + − +
+ −= +
∫
7. cos( )cos( )
dx
x a x b− −∫ =
(a) log Gcos ( x – a ) cos ( x – b ) G+c
(b) tan( )
logtan( )
x ac
x b
−+
−
(c) 1 cos( )
logsin( ) cos( )
x ac
a b x b
−+
− −
(d) 1 cos( )
logcos( ) cos( )
x ac
a b x b
−+
− −
SolutionSolutionSolutionSolution
18
( )
( ) ( )
( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
( )( ) ( )( )
( )( ) ( )
( )
( )
( )
( )
( )
( )
sin x b x a1I
sin a b cos x a cos x b
sin x b cos x a cos x b sin x a1
sin a b cos x a cos x b
1tan x b tan x a dx
sin a b
1log sec x b log sec x a
sin a b
sec x b1log c
sin a b sec x a
cos x a1log c
sin a b cos x b
− − − =
− − −
− − − − −=
− − −
= − − −−
= − − − −
−= +
− −
−= +
− −
∫
∫
∫
8. ( 1) 1
dx
x x+ −∫ =
(a) 1 1
2 tan2
xc−
−+
(b)
11 1tan
22
xc
− +
+
(c) 1 1
tan2
xc
− −
+
(d) 11 1
tan2 2
− +
+
xc
SolutionSolutionSolutionSolution
( )
2
2
22
1 1
put x 1 t dx 2 tdt
also x 1 t
2 t d t d tI 2
t 21 t 1 t
x 11 t2 tan c 2 tan c
2 2 2
− −
− = ∴ =
= +
= =++ +
− = × + = +
∫ ∫
9. 4 5sin
dx
x−∫
19
(a) 13 5 tan( / 2) 4
tan2 3
xc
− − +
(b)
( )
( )
2 tan / 2 41log
3 2 tan / 2 1
−
−
x
x
(c) log 5 4sinx c− + (d) 2
log 5 tan / 2 43
x c+ +
SolutionSolutionSolutionSolution
2 2
2
2
2
2 22
xput tan t
2
2 dt 2tdx sin x
1 t 1 t
2dt
1 t dtI 2
2t 4 4 t 10 t4 51 t
2 dt 1 dt 1 dt
5 t4 2 25 25 5 9t 1 t 1 t2 4 16 4 16
5 3t
1 1 4 4log3 5 32
2 t4 4 4
t 2 2 t 4 2 tan (x / 2) 41 1 1log log log
13 3 2 t 1 3t
2
=
∴ = =+ +
+= =
+ −−+
= = =
− + − − + − −
− −
= ×
× − +
− − −= = =
−−
∫ ∫
∫ ∫ ∫
2 tan (x / 2) 1−
10. cos
(1 sin )(2 sin )
x dx
x x− −∫
(a) 1 sin
log2 sin
xc
x
−+
− (b) log (1 sin )(2 sin )x x c− − +
(c) 2 sin
log1 sin
xc
x
−+
− (d)
(1 sin )log
(2 sin )
−+
−
xc
x
SoluSoluSoluSolutiontiontiontion
20
( ) ( )
( ) ( )
( ) ( )
put sin x t cos x dx dt
2 t 1 tdtI
1 t 2 t 1 t 2 t
1 1dt
1 t 2 t
log 1 t log 2 tI
1 1
2 t 2 sin xI log c log c
1 t 1 sin x
= ∴ =
− − −= =
− − − −
= −
− −
− −∴ = −
− −
− −= + = +
− −
∫ ∫
∫
11.
11 12 13
14 15
x x xdx
x x
+ +
+∫
(a) 14 15log x x c+ + (b) 2
1
1 2
xlog c
x x− +
+
(c) 2
1
1 2
xlog c
x x+ +
+ (d) 2
1
1− +
+
xlog c
x x
SolutionSolutionSolutionSolution
21
( )
( )
( )
( ) ( )
( )
( )
11 12 13 11 2
14 15 14
22
3 3
3
3
2
2
2
x x x x (1 x x )I dx dx
x x x (1 x )
1 x x1 x x
x 1 x x 1 x
1 x 1
x 1 xx 1 x
1 x xx dx dx
x 1 x
x 1 1dx
2 x 1 x
1log x log 1 x c
2 x
x1log c
1 x2 x
−
−
+ + + += =
+ +
+ ++ += =
+ +
+= +
++
+ −= +
+
= + −
− +
−= + − + +
−= + +
+
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∫
12. cosxe x dx∫
(a) 1
(sin 2cos )2
xe x x c+ + (b)
1(sin cos )
2
xe x x c− +
(c) 1
(cos sin )2
xe x x c− + (d)
1(sin cos )
2
xe x x c+ +
SolutionSolutionSolutionSolution
[ ]
[ ]
x
axa x
2 2
x
I e .cos x .dx
eusing e cos bx dx a cos bx b sin bx c
a b
eI cos x sin x c
1 1
=
= + ++
= + ++
∫
∫