Integrals Involving Trigonometric Substitution

7/29/2019 Integrals Involving Trigonometric Substitution

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Integrals Involving Trigonometric Substitution

I. Integrals of the Form 22 xa

Suppose that you have an integral of the form 22 xa where a is a positive

constant. If you make the substitution sinax , where22

, then

coscos)sin1(sin222222222 aaaaaxa because

0cos since22

. Also, dadx cos . Also, sinax

a

x

a

xarcsinsin .

x a

sya

s

sin andr = a

22 xa

a

sasaxasx

22

22222cos

.

Example: Find dxx 24 .

Let sin2x where 22

. Then cos242

x and dx

dcos2 . dddxx 22 cos4)cos2)(cos2(4 CC

cossin22cossin2

1

2

14 =

x 2

Cxxx

2

4

22

2arcsin2

2

=

2

4

2arcsin2

2xxx

+ C.

2

4 x

Lets check this answer.

2

4

2arcsin2

2xxx

dx

d

1


2/14

2

2

2

2

22424

24

42

2

2

1

2

1

21

12

x

x

xx

x

xx

x

2

2

2

2

222

442

)4(2

42

44

2

4

xx

x

x

xxx

and so it checks.

II. Integrals of the Form 22 xa

Suppose that you have an integral of the form 22 xa where a is a positive

constant. If you make the substitution tanax , where22

, then

secsec)tan1(tan222222222

aaaaaxa because

0sec since 22 . Also, dadx2sec . Also, tanax

a

x

a

xarctantan .

x22 xa

sya

stan and x = a

a

22222 sarrsa .

Example: Find dxx 42 .

Let tan2x where22

. Then sec242 x and dx =

d2

sec2 . dddxx 322 sec4)sec2)(sec2(4 CC

tansecln2tansec2tansecln

2

1tansec

2

14 =

C

xxxx

22

4ln2

22

42

22

x 42 x

Cxxxx

4ln22

4 22

.

2

2


3/14

Lets check this answer.

xxxxdx

d4ln24

2

1 22

41

42

2

4

124

2

1

4

2

2

1

2

2

22

2

2x

x

x

x

xxx

x

xx

42

82

42

44

4

4

4

12

2

4

2

2

2

22

2

2

2

2

x

x

x

xx

x

xx

xx

x

44

4 22

2

x

x

xand so it checks.

III. Integrals of the Form 22 ax

Suppose that you have an integral of the form 22 ax where a is a positive

constant. If you make the substitution secax , where 20

or 23

,

then tantan)1(secsec 222222222 aaaaaax because

0tan for2

0 or2

3 . Also, dadx tansec . Also,

a

xarc

a

xax secsecsec .

Considera

ssec where 0

20 s

x = a and r = s 22222 asysya . 22 ax x

Considerassec where 02

3 s

ax and 22222 asysyasr . a

Example: Find dxx

x

42

.

Let sec2x where2

0 or2

3 . Then tan242 x and

ddx tansec2 . Thus,

sec2

tansec2tan242

ddx

x

x

Cdd 2tan2)1(sec2tan2 22

Cx

arcx

2sec2

2

42

2

= 42 x x

3 2


4/14

Cx

arcx

2sec24

2 . This answer does check.

IV. Integrals of the Form ,, 2222 uaua and 22 au

A. Sometimes, you should make a usubstitution before you make a trigonometric

substitution.

Example: Find dxx

x

2

249

.

Let dxduxuxu 224 22 . Also, 392 aa .

dxx

x

2

249

= duu

udx

x

x

2

2

2

29

224

492 . Let sin3u

where22

. cos39 2 u and ddu cos3

d

ddu

u

u 222

2

cot2)sin3(

)cos3(cos32

92

Cd 2cot2)1(csc2 2

C

u

u

u

3arcsin2

92

2

u 3

C

x

x

x

3

2

arcsin22

49

2

2

=

Cx

x

x

3

2arcsin2

492

. 29 u

B. Sometimes, you have to complete the square before making a usubstitution or a

trigonometric substitution.

Example 1: Find dxxx 13841

2.

Complete the square for 1384 2 xx :

13)2(4138422 xxxx

= 413)12(42 xx

= 9)1(4 2 x

4


5/14

Let dxduxuxu 2)1(2)1(4 22 . Also, 392 aa .

Thus,

duudxxdxxx 9

1

2

1

9)1(4

1

1384

1222

Cx

Cu

3

)1(2arctan

6

1

3

arctan

3

1

2

1.

Example 2: Find dxxx 29615 .

Complete the square for 29615 xx :

1569961522 xxxx

153

29

2

xx

1159

1

3

2

9

2

xx 22)13(16)169(16 xxx

Let dxduxuxu 313)13( 22 . Also, 4162 aa .

Thus, dxxdxxx 3)13(1631

961522

duu 21631

. Let sin4u where22

. Then,

cos4162 u and ddu cos4 . duu

216

3

1=

Cdd cossin38

3

8cos

3

16)cos4)(cos4(

3

1 2

C

uuu

4

16

43

8

4arcsin

3

82

Cxxxx

6

9615)13(

4

13arcsin

3

82

. u 4

216 u

5


6/14

Practice Sheet for Trigonometric Substitution

(1) 22

425x

xdx =

(2) 10222

xx

xdx =

(3) 941

22xx

dx =

(4)

52

1

2xx

dx =

(5) 3

1

2215

1

xx

dx =

(6)

3

3

29 x dx =

(7) dxxx 4

1

23

(8) dxx 12

(9)

dxee

xx 2

1

(10) dxxx

x

267

6


7/14

(11)

duuau

2

22

(12)

dx

x

x2

2

259

(13)

dxxx 12

(14) dxxx

2

3

1

(15) dxx 1

12

(16)

dxx 2211

Solution Key for Trigonometric Substitution

(1) First, let dxduxuxu 224 22 22

425 x

xdx =

duu

udx

x

x

22

2

2

258

12

425

4

8

1. Next, make the trigonometric substitution

5

dduu cos5sin5 and cos525 2u u

225 u

7


8/14

Thus, duu

u

2

2

258

1 .sin8

25cos5

cos5

sin25

8

1 22

dd Using the

result

d

2sin cossin

2

1

2

1 , we get

16

25

4252

2

dx

x

x

5

2arcsin

16

25

5

25

516

25

5arcsin

16

25cossin

16

252

xC

uuuC

Cxxx

Cxx

22

4258

1

5

2arcsin

16

25

5

425

5

2

16

25.

(2) By long division, 1028

102

22

1102

102

1102 2222

2

xxxx

x

xx

x

xx

x

.

Thus,

dxxdxxxx

dxdxxx

x

9)1(

18

102

221

102222

2

Cx

xxx

2

1arctan

3

8102ln

2 .

(3) First, let dxduxuxu 224 22

94

1

22 xx

dx =

duuu

dxxx 9

122

944

12

2222. Next, make the trigonometric

substitution dxxxduu tansec3sec3 and tan392u

92 u u

3

8


9/14

Thus, duuu 9

12

22 dd cos9

2tansec3

)tan3(sec9

12

2

Cx

xCx

xCu

uC

9

942

94929

92sin

92

222

.

(4) First complete the square to get 4)1(52 22 xxx . Next, let 1xu dxdu 52

1

2xx

dx = duu 41

2. Make the trigonometric

substitution dduu 2sec2tan2 and sec242u

u 42 u

2

duu 41

2 = Cdd tanseclnsecsec2sec2

1 2=

CxxxCuuCuu

)1(52ln4ln22

4ln

22

2

.

(5) First, complete the square: 16)12(152215 222 xxxxxx

2)1(16 x . Next, let dxduxu 1

dxxx 22151

du

u 2161

. Make the trigonometric substitution dduu cos4sin4

9


10/14


11/14

3

3

29 x dx =

)1arcsin(2

91arcsin

2

9

2

9

3arcsin

2

93

3

2xxx

2

9

22

9

22

9

. [Note: dxx

3

3

2

9 represents the area of a semicircle

of radius 3,2

9)3(

2

1 2 .]

(7) Make the trigonometric substitution ddxx tansec2sec2 and

tan242x x

42 x

2

Thus, dxxx 4

1

23 dd

2

3cos

8

1tansec2

tan2sec8

1

Cx

xxarcC

2

2

8

4

2sec

16

1cossin

16

1

16

1 .

(8) Make the trigonometric substitution ddxx 2sectan and 12x

sec

x 12 x

1

Thus, dxx 12 tansec21

sec))(sec(sec32

dd

CxxxxCxx 1ln2

11

2

1tansecln

2

1 22 .

11


12/14

(9) Make the usubstitution duudxeedxedueuxxxx 22 11 .

Next, make the trigonometric substitution dduu cossin and

Cdduuu cossin21

2

1cos1cos1

222

Cuuu 212

1arcsin

2

1

Ceee xxx 212

1arcsin

2

1. u 1

21 u

(10) First, complete the square: 7)6(7667 222 xxxxxx 22

)3(1697)96( xxx . Next, let dxduxu 3 and

3ux

duuu

du

u

udx

xx

x

2221616

3

67

du

u

duuudu

u2

21

2

216

13)2(16

2

1

16

3

C

uuC

uu

4arcsin316

4arcsin3162

2

1 221

2

Cx

xx

4

3arcsin367

2 .

(11) Make the trigonometric substitution daduau 2sectan and

duuau

aau2

2222

sec

22

2

tan

)sec)(sec(

a

daa u 22 au

dd1sec

secsec

1sec

sec22

3

a

tanseclncotcscsectan

secsec

2dddd

12


13/14

Cu

auuauC

u

au

a

u

a

auC

22

22

2222

lnlncsc .

(12) Let dxduxuxu 33922

using Problem

Cx

xxxdu

u

udx

x

x

259

2593ln325

3)3(9

2593

22

2

2

2

2

.

(13) Make the trigonometric substitution ddxx tansecsec and

d

ddx

x

xx

22

2tan

sec

)tan)(sec(tan1tan1

Cd tan1sec2 x

Cxarcx sec12 . 12 x

1

(14) Make the trigonometric substitution ddxx cossin and 21 x

ddxx

xcos

cos

sin

1cos

3

2

3

Cd cos32

cossin3

1sin

23 x 1

Cxxx 222 13

21

3

1.

2

1 x

(15) Make the trigonometric substitution ddxx 2sectan and 12x

13


14/14

)(secsec1

1

1sec

2

2

ddx

x

Cd tanseclnsec x 12 x Cxx 1ln 2 .

1

(16) Make the trigonometric substitution dxxdxx 2sectan and 21 x

ddxx

4

2

22

2

sec

sec

)1(

1sec

Cd cossin2

1

2

1cos

2

x2

1 x

C

xx

xx

221

1

12

1arctan

2

1

Cx

xx

212

1arctan

2

1. 1

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Integrals Involving Trigonometric Substitution

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