03 04 trig substitution and partial fractions (1)

7/27/2019 03 04 Trig Substitution and Partial Fractions (1)

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Trigonometric Substitution PartialFraction

Trigonometric Substitution and Integration ofRational Functions by Partial Fractions

Mathematics 54Elementary Analysis 2

Institute of Mathematics

University of the Philippines-Diliman

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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises

Trigonometric Substitution

We use trigonometric substitution for integrands that contain

expression of the form:a2u2

a2+u2u2a2

Example

1

49x2 dxx

2

dx(x3)225

5

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Trigonometric Substitution

Case 1. Integrand contains the form

a2u2

Let u= asin, where

2,

2

. Note du= acosd.

a2

u2

=a2a2 sin2=a2(1 sin2)= a

cos2

= a|cos|

= acos

a2u2

ua

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Case 1.

a2u2. Let u= asin.

Example 1. Find 49x2 dx

x.

Let x

=7sin. Then dx

=7cos d.49x2

xdx=

7cos7sin

7cos d

49x2

x7

=

7cos2

sind =

7(1sin2)

sind

= 7(cscsin) d= 7 ln|csccot|+7cos+C

= 7 ln

7

x

49x2x

+

49x2+C.

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Trigonometric SubstitutionIntegrals involving

a2u2,

a2+u2 or

u2a2


a2+u2

Let u

=atan, where

2,

2 . Note du=asec2d.

a2+u2 =a2+a2 tan2

=

a2(1+ tan2)= a

sec2

= asec

a

u

a2+u2

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T i i S b i i P i lF i C C C E i


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Case 2.

a2+u2. Let u= atan.Example 2. Find

dx(x26x+25)3

.

We have: x26x+25= (x3)2+42, Letx

3

=4tan

dx

=4sec2 d

4

x

3(x3)2+42

dx

(x26x+25)3= dx

(x3)2+423

=

4sec2

(4sec)3d

= 116

cos d

= 116

sin+C

=1

16

x

3

x26x+25 +C6/28

T i t i S b tit ti P ti lF ti C 1 C 2 C 3 E i


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Trigonometric SubstitutionIntegrals involving

a2u2,

a2+u2 or

u2a2


u2a2

Let u

=asec, where

0,

2 , 3

2 . Note du=asec tand.

u2a2 =a2 sec2a2

=

a2(sec21)= a

tan2

= atan

a

u2a2

u

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Case 3.

u2a2. Let u= asec.Example 3. Find

e4

e2

ln2 x4xln x

dx.

Let ln x= 2sec. Then 1x

dx= 2sec tan d.

2

ln2 x4ln x

e

4

e2

ln

2

x4xln x

dx=

3

0

2tan2sec

2sec tan d

=

3

0

2tan

2sec2sec tan d

=

3

02tan2 d

= 2

3

0(sec21) d= 2(tan)

3

0

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Example on Area

Example 4. Evaluate 1

0

1x2 dx

From the definition of the definite integral, the expression

1

01x2 dxis the area of the region

.

By trigonometric substitution, let x= sin. Then dx= cosd10

1x2 dx =

2

0 cos cosd=

2

0

1

2(1+cos2)d

=1

2 (+1

4 sin2) 2

0 =1

2

2 +1

4 sin01

4 sin0 = 4

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Summary

a2

u2 Let u

=asin

a2+u2 Let u= atan

u2a2 Let u= asec

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Exercises

Evaluate the following integrals.

1

dx

x2+2x152

16

e2x

ex dx

3

42

x24

xdx

4 x3 dx

1x25

dx

25+x23

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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises


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g

Introduction

Recall.

A function his a rational functionifh(x)= f(x)g(x)

, where fand g are

polynomial functions.

Consider

1x2+5x+6 dx.

Solvable using math 53:

1

x2

+5x+6dx

=1

x+

522 14

dx

=

1

2 1

2

ln x+3

x+2 +C

Note:1

x2+5x+6 dx=

1

x+3 1

x+2

dx= ln

x+3x+2

+C

This integral will be easier to evaluate.12/28



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g

Goal.

To decompose a rational expression as a sum of two or more

simpler quotients, called partial fractions.

Remark.

1 We will consider rational functions h(x)= f(x)g(x)

with

deg(f(x))< deg(g(x)) (fand g are polynomial functions). Ifdeg(f(x)) deg(g(x)), then we divide first the numerator by thedenominator such that

h(x)

=q(x)

+

r(x)

g(x)

,

where deg(r(x))< deg(g(x)).2 Any polynomial with real number coefficients can be

expressed as a product of linear and quadratic polynomials.

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Case 1. Denominator of the rational function only has

distinct linear factors.

Ifg(x)= (a1x+b1)(a2x+b2) (anx+bn), where all factors aredistinct, then

p(x)

q(x)= A1

a1x+b1+ A2

a2x+b2+ + An

anx+bn.

Example 1. Find x dx

x25x+6 .

x

x25x+6 =x

(x3)(x2) =A

x3 +B

x2 =A(x2)+B(x3)

(x3)(x2) .

We have:

x=A(x2)+B(x3).

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Case 1. Factors of the denominator are linear.

Example 2. Find 2x34x215x+5

x2

2x

8

dx.

Since the degree of the numerator is greater than the denominator,

we divide first the numerator by the denominator. We have

2x34x215x+5

x22x8 =

2x

+

x+5

(x4)(x+2) =2x

+

A

x4 +

B

x+2.

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3 2


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2x34x215x+5x22x8 = 2x+ Ax4 + Bx+2

x+

5

(x4)(x+2) =A

x4 +B

x+2 x+5=A(x+2)+B(x4)

Substitute2 for x: 3=B(6)B= 12

Substitute 4 for x: 9=A(6)A=3

2

Hence,

2x34x215x+5

x22x8 dx =

2x+ x+5(x4)(x+2)

dx

= 2x22+

32(x4)

12(x+2)

dx

= x2+ 32

ln |x4| 12

ln |x+2|+C.

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Case 2. Factors of denominator are all linear but some are

repeated.

If (ax+b)n, n> 1 is a factor of the denominator, then the partialfractions corresponding to this factor are

A1

ax

+b+ A2

(ax

+b)2

+ + An(ax

+b)n

.

Example 3. Find

(3x2) dx

2x

3

x2

.

3x22x3x2 =

3x2x2(2x1) =

A

x+ B

x2+ C

2x1 .

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3x22x3x2 = Ax+ Bx2 + C2x1

Multiplying both sides byx2(2x1)

3x2=Ax(2x1)+B(2x1)+Cx2

let x=

0

B=

2

let x= 12

C=2let x= 1 A= 1

(3x2) dx

2x3x2 =

1

x+ 2

x2 2

2x1

dx

= ln |x|+ 2x1

1 ln |2x1|+C

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More Examples

Example 4. Find

(x5) dx

(x+1)2(x2) .

x5

(x+1)2

(x2) =

A

x+1 +

B

(x+1)2

+

C

x2.

We have x5=A(x+1)(x2)+B(x2)+C(x+1)2.Substitute1 for x: 6=B(3)B= 2

Substitute 2 for x: 3=C(3)2 C= 13

Substitute 0 for x: 5=A(1)(2)+2(2)+ 1

3 (1)

2

A=1

3x5

(x+1)2(x2) dx =

1

3(x+1) +2

(x+1)2 1

3(x2)

dx

=

1

3

ln

|x

+1

|+

2(x+1)1

1

1

3

ln

|x

2

|+C

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Case 3. Factors of the denominator are linear and

quadratic and none of the quadratic factors is repeated.

Ifax

2

+bx+ cwhere a= 0 is a factor of the denominator that is notrepeated , then the corresponding partial fraction to this factor isAx+B

ax2+bx+c.

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Case 3. Factors of the denominator are linear and

quadratic and none of the quadratic factors is repeated.

Example 5. Find 5x2+3x2

x31 dx.

5x2+3x2

x31 dx=

5x2+3x2(x1)(x2+x+1) dx=

A

x1 +Bx+C

x2+x+1

dx

Consider 5x2

+3x

2=

A(x2

+x+

1)+

(Bx+

C)(x

1).

Substitute x= 1: 6=A(3)A= 2Substitute x= 0: 2= 2(1)+C(1)C= 4

Substitute x=1: 0= 2(1)+ (B+4)(2)B= 3

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5x2+3x 2 2 3x+4


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5x2+3x2x31 = 2x1 + 3x+4x2+x+1

5x2+3x2

x31 dx

= 2

x

1dx+

3x+4x2

+x+

1dx

=

2

x1 dx+ 3

2(2x+1)

x2+x+1 dx+ 5

2

x2+x+1 dx

=2 ln

|x

1

|+3

2

ln

|x2

+x

+1

|+5

2 1

3/2tan1

x+ 12

3/2+C

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d i f f h d i


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Case 4. Some quadratic factors of the denominator are

repeated.

If (ax2+bx+c)n, n> 1 is a factor of the denominator , then the

corresponding partial fractions are

A1x+b1ax2+bx+c+

A2x+b2(ax2+bx+ c)2 + +

Anx+bn(ax2+bx+c)n

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C 4 S d i f f h d i


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Case 4. Some quadratic factors of the denominator are

repeated.

Example 6. Find

x3+1

(x2

+4)2

dx.

x3+1

(x2+4)2 dx=

Ax+Bx2+4 dx+

Cx+D

(x2+4)2 dx

x3

+1=

(Ax+

B)(x2

+4)+

(Cx+

D)=

Ax3

+Bx2

+(4A

+C)x

+D

By comparing coefficients, we have A= 1, B= 0, D= 1.Also, ifx= 1 then 2=A+B+ (4A+C)+D.C=4

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C ti ti


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Continuation...

x3+1

(x2+4)2 dx =

x

x2+4 dx+

14x(x2+4)2 dx

=1

2 2x

x2+4 dx+ 14x(x2+4)2 dx

= 12

2x

x2+4 dx2

2x

(x2+4)2 dx+

dx

(x2+4)2

=1

2 ln |x2

+4|2(x2

+4)1

1 + dx(x2+4)2 (trig.sub.)

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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercisesx3+1 d 1 l | 2+4|+ 2 +

dx


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x +1

(x2+4)2 dx= 12 ln |x2+4|+ 2x2+4 +

dx(x2+4)2

Let x= 2tan dx= 2sec2 d

2

x

x2+22

dx

(x2+4)2 =

2sec2 d

16sec4

= 18

cos2 d

=1

8 1+cos22 d

= 116+ 1

32sin2+C

=

1

16

+

1

32 2sincos

+C

Hence,x3+1

(x2+4)2 dx=1

2ln |x2+4|+ 2

x2+4+1

16tan1

x2

+ 1

16

xx2+4

2x2+4

+C.

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E ercises


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Exercises

Decompose the following into partial fractions.

1(4x3)

x22x32 1

x3(x1)3

2x+1x2(x2+2)

4 x3

2x(x2+1)2

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03 04 trig substitution and partial fractions (1)

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