Inverse substitution rule Inverse Substitution Rule If and

is differentiable and invertible. Then

( ( )) ( ) ( )f g t g t dt F t C ( )x g t

1( ) ( ( )) ( ) ( ) ( ( )) .f x dx f g t g t dt F t C F g x C

Example: trigonometric substitutions Ex. Evaluate

Sol. Let then

From we derive

2 2 ( 0).a x dx a sin ,x a t cos .dx a tdt

2 2 2 2cos cos cosa x dx a t a tdt a tdt 2 2 sin 2

(1 cos 2 ) ( )2 2 2

a a tt dt t C

sin ,x a t2 2

arcsin , cos .x a x

t ta a

2 2 22 2 arcsin .

2 2

a x x a xa x dx C

a

Example: trigonometric substitutions Ex. Evaluate

Sol. Let then

From we derive

2 2( 0).

dxa

x a

tan ,x a t 2sec .dx a tdt1 1 sin

ln | |2 1 sin

tC

t

tan ,x a t2 2

sin .x

tx a

2

2 2

secsec

sec

dx a tdt tdt

a tx a

2 2 2 2 2

22 2

1 ( )ln ln .

2

dx x a x x a xC C

a ax a

Example Ex. Evaluate

Sol.

2

2

9.

xdx

x

22 2

2 2 2

9 (3sin )9 cos(3sin )

(3sin ) sin

tx tdx d t dt

x t t

22 9

(csc 1) cot arcsin .3

x xt dt t t C C

x

Example Ex. Evaluate substitution

Ex. Find substitution

Ex. Find

substitution

2 2

1.

4dx

x x 2 tanx t

2 2

1.

4dx

x x 2secx t

2

1.

3 2dx

x x x 2 23 2 4 ( 1)x x x 1 2sinx t

Example: reciprocal substitution Ex. Evaluate

Sol. Let then

2.

3 2 1

dx

x x x

1,x

t

2

1.dx dt

t

2 2 2 23 2 1 3 2 2 ( 1)

dx dt dt

x x x t t t

2

1 1 1arcsin arcsin .

2 2 211 ( )

2

dt t xC C

xt

Example: rational substitution Ex. Evaluate

Sol. Let then

.1

xdx

x2 ,x t

2 3

2 21 1

t ttdt dt

t t

3

2 21 22[ ( 1) ] 2 2ln(1 )

1 3

t

t t dt dt t t t Ct

3 1 12

1

t

dtt

322 2ln(1 ) .

3

xx x x C

2 .dx tdt

1

xdx

x

Inverse substitution for definite integral

The Inverse Substitution Rule for definite integral: If

x=g(t) is differentiable, invertible and, when x is in between

a and b, t is in between and Then1( )g a 1( ).g b

( ) ( ( )) ( ) .b

af x dx f g t g t dt

Example Ex. Evaluate

Sol. Let then and when x

changes from a to 2a, t changes from 0 to

2 22

4( 0).

a

a

x adx a

x

sec ,x a t sec tan ,dx a t t/ 3.

2 22

34 4 40

tansec tan

sec

a

a

x a a tdx a t tdt

x a t

32 33

2 2 200

1 1 3sin cos sin .

3 8t tdt t

a a a

Example Ex. Evaluate

Sol. Since let

Then and when x changes from 2 to 3, t changes from 0 to

3

2 32

1.

(4 )

xdx

x x

2 2(4 ) 4 ( 2) ,x x x 2 2sin .x t 2cos ,dx t

/ 6.3

6 63 22 32 0 0

1 3 2sin 3 2sin2cos

(2cos ) 4cos(4 )

x t tdx tdt dt

t tx x

6

0

3 1 3 3 1 7 3 6tan ( ) .

4 2cos 4 3 2 12t

t

Example: application of substitution Ex. Find

Sol. Let then

2

0

sin.

sin cos

xdx

x x

02 2

0 02

sin cos cos

sin cos cos sin sin cos

x t tdx dt dt

x x t t t t

,2

x t

2 2 2

0 0 0

sin 1 cos sin( ) .

sin cos 2 sin cos sin cos 4

x t xdx dt dx

x x t t x x

Example: application of substitution Ex. Find the definite integral

Sol.

By substitution

22

2

sin.

1 x

xdx

e

2 2 20

2 2

022

sin sin sin.

1 1 1x x x

x x xdx dx dx

e e e

2 2 2 20 0

2 2

0 02 2

sin sin sin sin.

1 1 1 1

x

x t x x

x t x e xdx dt dx dx

e e e e

,x t

2 2 222 2 2 2

0 0 02

sin sin sinsin .

1 1 1 4

x

x x x

x e x xdx dx dx xdx

e e e

Integration of rational functions Any rational function can be integrated by the following

two steps: a). express it as a sum of simpler fractions by partial fraction technique; b. integrate each partial fraction using the integration techniques we have learned.

For example, since

we have

2

5 5 2 1

2 ( 1)( 2) 1 2

x x

x x x x x x

2

5 2 1( ) 2 ln | 1| ln | 2 | .

2 1 2

xdx dx x x C

x x x x

Technique for partial fraction Take any rational function where P and Q are

polynomials. If the degree of P is less than the degree of Q, we call f a

proper fraction. If f is improper, that is, degree of P greater than or equal to

degree of Q, then (by long division) we must have

where S and R are also polynomials and degree of R less than degree of Q.

( )( ) ,

( )

P xf x

Q x

( ) ( )( ) ( ) ,

( ) ( )

P x R xf x S x

Q x Q x

Technique for partial fraction For example, by long division, we can derive

For the above reason, we only need to consider the properrational functions. The next step is to factor the denominator Q(x). It can beshown that any polynomial can be factored as a product oflinear factors (in the form ax+b) and irreducible quadraticfactors (in the form ). Forexample,

32 2

2 .1 1

x xx x

x x

2 2 with 4 0ax bx c b ac 4 2( ) -16=(x-2)(x+2)(x 4).Q x x

4 2 2 2+1 ( 1)( 1)x x x x x x

Technique for partial fraction The third step is to express the proper rational function

R(x)/Q(x) as a sum of partial fractions of the form

These two kind of rational functions can be integrated as

2 or .

( ) ( )i j

A Ax B

ax b ax bx c

11 1 1 1ln | | , ( )

( ) (1 )k

kdx ax b C dx ax b C

ax b a ax b a k

2

2 2 2

( ).

( ) ( ) [1 ( ) ]k k k

Ax B d ax bx c Fdx C D dx

ax bx c ax bx c x E

Example Ex. Find

Sol.

2

1.

( 1)n nI dx

x

2 2 2

1

( 1) ( 1) ( 1)n n n n

dx xI xd

x x x

2

2 2 12

( 1) ( 1)n n

x x dxn

x x

2

2 2 1

( 1 1)2

( 1) ( 1)n n

x x dxn

x x

122 2 .

( 1) n nn

xnI nI

x

1 2

2 1.

2 2 ( 1)n n n

n xI I

n n x

1 arctan .I x C

2 2

arctan.

2 2( 1)

x xI C

x

Technique for partial fraction From the above analysis, we see that how to split a rational

function into partial fractions is the key step to integrate the rational function.

When Q(x) contains factor the partial fractions contain

When Q(x) contains irreducible factor the partial fractions contain

( ) ,kx a

1 22

.( ) ( )

kk

AA A

x a x a x a

2( ) ,kx px q

1 1 2 22 2 2 2

.( ) ( )

k kk

B x DB x D B x D

x px q x px q x px q

Example Ex. Find

Sol. Since the partial fraction has the form

Expanding the right side and comparing with the left side,

2

3 2

2 1.

2 3 2

x xdx

x x x

3 22 3 2 ( 2)(2 1),x x x x x x

2

3 2

2 1.

2 3 2 2 2 1

x x A B C

x x x x x x

2 2 1 ( 2)(2 1) (2 1) ( 2). x x A x x Bx x Cx x2(2 2 ) (3 2 ) 2 A B C x A B C x A

1 1 1, ,

2 5 10 A B C

Example Ex. Find Sol.

5 3 2

4 2

2 7.

2 2

x x x x

I dxx x x

5 3 2

4 2 4 2

2 7 5

2 2 2 2

x x x x x

xx x x x x x

4 2 2 22 2 ( 1) ( 2 2) x x x x x x

1 24 2 2 2

5

2 2 1 ( 1) 2 2

A Ax Bx D

x x x x x x x

1 2

1 1 8, 1, ,

5 5 5 A A B D

221 1 1 7

ln | 1| ln( 2 2) arctan( 1) .2 5 1 10 5

xI x x x x C

x

Remark There are two methods to find the coefficients in the

partial fractions. One method is comparing the corresponding coefficients of polynomials on both sides; the other is taking some special values of x in the identity.

For instance, in the last example, we have2 2 2

1 25 ( 1)( 2 2) ( 2 2) ( )( 2 1) x A x x x A x x Bx D x x

2 21 5 5 1 x A A1 8

1 5 5 ( 3 ) (7 4 ) ,5 5

x i i B D B D i B D

1 2 1

10 0 2 2

5x A A D A

Example Ex. Find Sol.

4 3 2

2 2 2

3.

( 1)( 1)

x x x x

I dxx x

4 3 21 2 1 1 2 2

2 2 2 2 2 2

3

( 1)( 1) 1 1 1 ( 1)

A A B x D B x Dx x x x

x x x x x x

1 2 1 1 2 2

1 1, , 1, 1, 1, 1

2 2 A A B D B D

22 2

1 1 1 1ln | 1| ln | 1| ln( 1) arctan

2 2 2 ( 1)

xI x x x x dx

x

2

2 2

1 1 1 1ln | | arctan .

2 1 2 2( 1)

x xx C

x x

Homework 18 Section 7.2: 18, 24, 42, 44

Section 7.3: 5, 6, 24, 27