Section 7.5 Strategy for Integration

(1)

Section 7.5 Strategy for Integration

8. Three integrals are given that, although they look similar, may require different techniques of integration. Evaluate the integrals.

(a) Z

e^x√

e^x− 1dx (b)

Z e^x

√1 − e^2xdx (c)

Z 1

√e^x− 1dx Solution:

SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 731

6. (a) Let  = ², so that  = 2  ⇒ ¹2 =  . Thus,



 cos ² =1 2



cos   = 1

2sin  +  = 1

2sin ²+ .

(b)

 cos²  = ¹₂

(1 + cos 2)  = ¹₂

  +¹₂

 cos 2  =¹₄²+¹₂ cos 2 .

The remaining integral can be evaluated using integration by parts with  = ,  = cos 2  ⇒

 = ,  = ¹₂sin 2. Thus,

 cos²  = ¹₄²+¹₂1

2 sin 2 − 1

2sin 2 

= ¹₄²+¹₂1

2 sin 2 +¹₄cos 2 + 

= ¹₄²+¹₄ sin 2 +¹₈cos 2 + 

(c) First, use integration by parts with  = ²,  = cos   ⇒  = 2 ,  = sin . This gives

 =

²cos   = ²sin  − 2 sin  . Next, use integration by parts for the remaining integral with  = 2,  = sin   ⇒  = 2 ,  = −cos . Thus,

 = ²sin  − (−2 cos  +

2 cos  ) = ²sin  + 2 cos  − 2 sin  + .

7. (a) Let  = ³, so that  = 3² ⇒ ¹3 = ². Thus,



²^³ =1 3



^ = 1

3^+  = 1

3^³+ . (b) First, use integration by parts with  = ²,  = ^ ⇒  = 2 ,  = ^. This gives

 =

²^ = ²^−

2^. Next, use integration by parts for the remaining integral with  = 2,

 = ^ ⇒  = 2 ,  = ^. Thus,  = ²^−

2^−

2^

= ²^− 2^+ 2^+ .

(c) Let  = ², so that  = 2 . Thus,

³^² =

²^²  = ¹₂

^. Now use integration by parts with  = ,  = ^ ⇒  = ,  = ^. This gives

³^² = ¹₂

^−

^

= ¹₂^−¹2^+  = ¹₂²^²− ¹2^²+ .

8. (a) Let  = ^− 1, so that  = ^. Thus,

^√

^− 1  =

^12 = ²₃^32+  = ²₃(^− 1)^32+ .

(b) Let  = ^, so that  = ^. Thus, ^

√1 − ^2 =

 1

√1 − ² = sin⁻¹ +  = sin⁻¹(^) + .

(c) Let  =√

^− 1, so that ²= ^− 1 ⇒ 2  = ^, and 2 

²+ 1 = . Then

 1

√^− 1 =

 1

 2 

²+ 1 = 2

 1

²+ 1 = 2 tan⁻¹ +  = 2 tan⁻¹√

^− 1 + .

9. Let  = 1 − sin . Then  = − cos   ⇒

 cos  1 − sin  =

 1

(−) = − ln || +  = − ln |1 − sin | +  = − ln(1 − sin ) + .

10. Let  = 3 + 1. Then  = 3  ⇒

 1 0

(3 + 1)^√² =

 4 1

^√²

1 3



= 1 3

 1

√2 + 1^√²⁺¹

4

1

= 1

3√

2 + 1

4^√²⁺¹− 1 .

27. Evaluate the integral.

Z

e^x+e^xdx.

Solution:

SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 673

18. Let  =√. Then  = 1 2√

 ⇒

 4 1

^√^

√  =

 2 1

^(2 ) = 2

^2

1= 2(²− ).

19. Let  = ^. Then

^+^ =

^^^ =

^ = ^+  = ^^+ .

20. Since ²is a constant,

² = ² + .

21. Let  =√

, so that ²= and 2  = . Then

arctan√

  =

arctan  (2 ) = . Now use parts with

 = arctan ,  = 2  ⇒  = 1

1 + ²,  = ². Thus,

 = ²arctan  −

 ²

1 + ² = ²arctan  −

 

1 − 1 1 + ²



 = ²arctan  −  + arctan  + 

=  arctan√

 −√

 + arctan√

 +  

or ( + 1) arctan√

 −√

 + 

22. Let  = 1 + (ln )², so that  = 2 ln 

 . Then

 ln 



1 + (ln )² = 1 2

 1

√ =1 2

 2√



+  =

1 + (ln )²+ .

23. 3²− 2

²− 2 − 8 = 3 + 6 + 22

( − 4)( + 2) = 3 + 

 − 4+ 

 + 2 ⇒ 6 + 22 = ( + 2) + ( − 4). Setting

 = 4gives 46 = 6, so  = ²³₃. Setting  = −2 gives 10 = −6, so  = −⁵₃. Now

 3²− 2

²− 2 − 8 =

 

3 + 233

 − 4 − 53

 + 2



 = 3 + ²³₃ ln | − 4| −⁵3ln | + 2| + .

24. 

(1 + tan )²sec   =

(1 + 2 tan  + tan²) sec  

=

[sec  + 2 sec  tan  + (sec² − 1) sec ]  =

(2 sec  tan  + sec³) 

= 2 sec  + ¹₂(sec  tan  + ln |sec  + tan |) +  [by Example 7.2.8]

25.  1 0

1 + 12

1 + 3  =

 1 0

(12 + 4) − 3 3 + 1  =

 1 0



4 − 3 3 + 1



 =

4 − ln |3 + 1|1

0= (4 − ln 4) − (0 − 0) = 4 − ln 4

26. 3²+ 1

³+ ²+  + 1 = 3²+ 1

(²+ 1)( + 1) = 

 + 1 + + 

²+ 1 . Multiply by ( + 1)(²+ 1)to get

3²+ 1 = (²+ 1) + ( + )( + 1) ⇔ 3²+ 1 = ( + )²+ ( + ) + ( + ). Substituting −1 for  gives 4 = 2 ⇔  = 2. Equating coefﬁcients of ²gives 3 =  +  = 2 +  ⇔  = 1. Equating coefﬁcients of  gives 0 =  +  = 1 +  ⇔  = −1. Thus,

 1 0

3²+ 1

³+ ²+  + 1 =

 1 0

 2

 + 1+  − 1

²+ 1



 =

 1 0

 2

 + 1+ 

²+ 1− 1

²+ 1





=

2 ln | + 1| +¹2ln(²+ 1) − tan⁻¹1

0= (2 ln 2 + ¹₂ln 2 − ^4) − (0 + 0 − 0)

= ⁵₂ln 2 − ^4

Z 1 + sin x 1 + cos xdx.

Solution:

] 1 + sin { 1 + cos {g{ =

] (1 + sin {)(1 3 cos {) (1 + cos {)(1 3 cos {)g{ =

] 1 3 cos { + sin { 3 sin { cos {

sin²{ g{

=]

csc²{ 3 cos {

sin²{+ csc { 3cos { sin {

g{

= 3 cot { +V 1

sin {+ ln |csc { 3 cot {| 3 ln |sin {| + F >E\ ([HUFLVH @

7KH DQVZHU FDQ EH ZULWWHQ DV1 3 cos {

sin { 3 ln(1 + cos {) + F

Z x²

x⁶+ 3x³+ 2dx.

Solution:

680 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

64.Let  = √ + 1, so that  = ( − 1)²and  = 2( − 1) . Then

 1

√ + 1 =

 2( − 1) 

√ =



(2^12− 2^−12)  = 4

3^32− 4^12+  =4 3

√ + 132

− 4√ + 1 + .

65.Let  = cos², so that  = 2 cos  (− sin ) . Then

 sin 2

1 + cos⁴ =

 2 sin  cos  1 + (cos²)²  =

 1

1 + ²(−) = − tan⁻¹ +  = − tan⁻¹(cos²) + .

66.Let  = tan . Then

3

4

ln(tan ) 

sin  cos  =

3

4

ln(tan )

tan  sec²  =

 ^√3 1

ln 

  =1

2(ln )²^√³

1 =¹₂ ln√

32

=¹₈(ln 3)².

67. 

√ + 1 +√

 =

  1

√ + 1 +√

·

√ + 1 −√

√ + 1 −√





 = 

 + 1 −√





=²₃

( + 1)^32− ^32 + 

68.

 ²

⁶+ 3³+ 2 =

 ²

(³+ 1)(³+ 2) =

 1

3

( + 1)( + 2)

 = ³

 = 3²

 .

Now 1

( + 1)( + 2) = 

 + 1+ 

 + 2 ⇒ 1 = ( + 2) + ( + 1). Setting  = −2 gives  = −1. Setting  = −1 gives  = 1. Thus,

1 3

 

( + 1)( + 2)= 1 3

  1

 + 1− 1

 + 2



 = 1

3ln | + 1| −1

3ln | + 2| + 

= ¹₃ln³+ 1

 −¹3ln³+ 2 + 

69.Let  = tan , so that  = sec² ,  =√

3 ⇒  =^3, and  = 1 ⇒  =^4. Then

^√3 1

√1 + ²

²  =

3

4

sec 

tan² sec²  =

 3

4

sec  (tan² + 1) tan²  =

 3

4

sec  tan²

tan² + sec  tan²





=

3

4

(sec  + csc  cot )  =

ln |sec  + tan | − csc 3

4

=

ln2 +√ 3

 −^√²₃

−

ln√2 + 1

 −√ 2

=√

2 −^√²₃+ ln 2 +√

3

− ln 1 +√

2

70.Let  = ^. Then  = ln ,  =  ⇒

 

1 + 2^− ^−=

 

1 + 2 − 1 =

 

2²+  − 1=

  23

2 − 1− 13

 + 1





=¹₃ln|2 − 1| −¹3ln | + 1| +  =¹3ln|(2^− 1)(^+ 1)| + 

1

(2)

93. Evaluate the integral Z π/6

0

√1 + sin 2θdθ.

Solution:

744 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 92.

 1

(sin  + cos )²  =

 1

sin² + 2 sin  cos  + cos² =

 1

cos²

sin²

cos²+2 sin  cos  + 1

 

=

 1

cos² (tan² + 2 tan  + 1) =

 1

cos² (tan  + 1)² =

 sec² (tan  + 1)² 

=

 1

² 

 = tan  + 1

 = sec² 



= −1

+  = − 1

tan  + 1+ 

93.

 6 0

√1 + sin 2  =

 6 0

sin² + cos²

+ 2 sin  cos   =

 6 0



(sin  + cos )²

=

 6 0

|sin  + cos |  =

 6 0

(sin  + cos ) 

 since integrand is positive on [0 6]



=

−cos  + sin 6

0 =



−

√3 2 +1

2



− (−1 + 0) =3 −√ 3 2 Alternate solution:

 6 0

√1 + sin 2  =

 6 0

√1 + sin 2 ·

√1 − sin 2

√1 − sin 2 =

 6 0

1 − sin²2

√1 − sin 2 

=

 6 0

√cos² 2

√1 − sin 2 =

 6 0

|cos 2|

√1 − sin 2 =

6 0

cos 2

√1 − sin 2

= −1 2

 1−√ 32

1

^−12 [^{ = 1}− sin 2  = −2 cos 2 ]

= −¹2



2^121−√ 32

1 = 1 −

 1 −√

32

94. (a)

 2 1

^

  =

 ln 2 0

^^

^ ^

 = ^

 = ^



=

 ln 2 0

^^ =  (ln 2)

(b)

 3 2

1 ln  =

 ln 3 ln 2

1

(^)



 = ln 

 =1





=

 ln ln 3 ln ln 2

^^

^ ^

  = ^

 = ^



=

 0 ln ln 2

^^ +

 ln ln 3 0

^^ [note that ln ln 2  0]

=

 ln ln 3 0

^^ −

 ln ln 2 0

^^ =  (ln ln 3) −  (ln ln 2)

Another method: Substitute  = ^^in the original integral.

95. The function  = 2^²does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.

(2²+ 1)^² =

2²^² +

^² =

 2^²

 +

^²

= ^²−

^² +

^²

  = ,

 = 

= 2^2,

= ^2



= ^²+ 

2