7 TECHNIQUES OF INTEGRATION

[Pages:209]7 TECHNIQUES OF

INTEGRATION

7.1 Integration by Parts

Preliminary Questions

1. Which derivative rule is used to derive the Integration by Parts formula? solution The Integration by Parts formula is derived from the Product Rule.

2. For each of the following integrals, state whether substitution or Integration by Parts should be used:

x cos(x2) dx,

x cos x dx,

x2ex dx,

xex2 dx

solution (a) x cos(x2) dx: use the substitution u = x2. (b) x cos x dx: use Integration by Parts. (c) x2ex dx; use Integration by Parts. (d) xex2 dx; use the substitution u = x2.

3. Why is u = cos x, v = x a poor choice for evaluating x cos x dx?

solution

Transforming v

=

x

into v

=

1 2

x2

increases the power of x

and makes

the new integral harder

than the

original.

Exercises

In Exercises 1?6, evaluate the integral using the Integration by Parts formula with the given choice of u and v .

1. x sin x dx; u = x, v = sin x solution Using the given choice of u and v results in

u = x v = - cos x u = 1 v = sin x Using Integration by Parts,

x sin x dx = x(- cos x) - (1)(- cos x) dx = -x cos x + cos x dx = -x cos x + sin x + C.

2. xe2x dx; u = x, v = e2x

solution Using u = x and v = e2x gives us

u=x

v

=

1 2

e2x

u = 1 v = e2x

Integration by Parts gives us

xe2x dx = x 1 e2x - (1) 1 e2x dx = 1 xe2x - 1 1 e2x + C = 1 e2x (2x - 1) + C.

2

2

2

22

4

807

March 30, 2011

808 C H A P T E R 7 TECHNIQUES OF INTEGRATION

3. (2x + 9)ex dx; u = 2x + 9, v = ex

solution Using u = 2x + 9 and v = ex gives us

u = 2x + 9 v = ex

u =2

v = ex

Integration by Parts gives us

(2x + 9)ex dx = (2x + 9)ex - 2ex dx = (2x + 9)ex - 2ex + C = ex (2x + 7) + C.

4. x cos 4x dx; u = x, v = cos 4x

solution Using u = x and v = cos 4x gives us

u=x

v

=

1 4

sin 4x

u = 1 v = cos 4x

Integration by Parts gives us

x cos 4x dx = 1 x sin 4x - (1) 1 sin 4x dx = 1 x sin 4x - 1 - 1 cos 4x + C

4

4

4

44

= 1 x sin 4x + 1 cos 4x + C.

4

16

5. x3 ln x dx; u = ln x, v = x3

solution Using u = ln x and v = x3 gives us

u = ln x

v

=

1 4

x4

u

=

1 x

v = x3

Integration by Parts gives us

x3 ln x dx = (ln x) 1 x4 - 4

1 x

1x4 dx 4

= 1 x4 ln x - 1 x3 dx = 1 x4 ln x - 1 x4 + C = x4 (4 ln x - 1) + C.

4

4

4

16

16

6. tan-1 x dx; u = tan-1 x, v = 1

solution Using u = tan-1 x and v = 1 gives us

Integration by Parts gives us

u = tan-1 x

u

=

1 x2 + 1

v=x v =1

tan-1 x dx = x tan-1 x -

1 x dx. x2 + 1

For the integral on the right we'll use the substitution w = x2 + 1, dw = 2x dx. Then we have

tan-1 x dx = x tan-1 x - 1 2

1 x2 + 1

2x dx = x tan-1 x - 1 2

dw w

= x tan-1 x - 1 ln |w| + C = x tan-1 x - 1 ln |x2 + 1| + C.

2

2

March 30, 2011

S E C T I O N 7.1 Integration by Parts 809

In Exercises 7?36, evaluate using Integration by Parts.

7. (4x - 3)e-x dx

solution Let u = 4x - 3 and v = e-x . Then we have

u = 4x - 3 v = -e-x

u =4

v = e-x

Using Integration by Parts, we get

(4x - 3)e-x dx = (4x - 3)(-e-x ) - (4)(-e-x ) dx

= -e-x (4x - 3) + 4 e-x dx = -e-x (4x - 3) - 4e-x + C = -e-x (4x + 1) + C.

8. (2x + 1)ex dx

solution Let u = 2x + 1 and v = e-x . Then we have

u = 2x + 1 v = -e-x

u =2

v = e-x

Using Integration by Parts, we get

(2x + 1) e-x dx = (2x + 1)(-e-x ) - (2)(-e-x ) dx

= -(2x + 1)e-x + 2 e-x dx = -(2x + 1)e-x - 2e-x + C = -e-x (2x + 3) + C.

9. x e5x+2 dx solution Let u = x and v = e5x+2. Then we have

u=x u =1

v = 1 e5x+2 5

v = e5x+2

Using Integration by Parts, we get

xe5x+2 dx = x 1 e5x+2 - (1) 1 e5x+2 dx = 1 xe5x+2 - 1

5

5

5

5

= 1 xe5x+2 - 1 e5x+2 + C = x - 1 e5x+2 + C

5

25

5 25

e5x+2 dx

10. x2ex dx solution Let u = x2 and v = ex . Then we have

u = x2 v = ex u = 2x v = ex Using Integration by Parts, we get x2 ex dx = x2ex - 2 xex dx.

We must apply Integration by Parts again to evaluate xex dx. Taking u = x and v = ex , we get

xex dx = xex - (1)ex dx = xex - ex + C. Plugging this into the original equation gives us

x2 ex dx = x2ex - 2 xex - ex + C = ex (x2 - 2x + 2) + C.

March 30, 2011

810 C H A P T E R 7 TECHNIQUES OF INTEGRATION

11. x cos 2x dx

solution Let u = x and v = cos 2x. Then we have

u=x

v

=

1 2

sin 2x

u = 1 v = cos 2x

Using Integration by Parts, we get

x cos 2x dx = x 1 sin 2x - (1) 1 sin 2x dx

2

2

= 1 x sin 2x - 1 sin 2x dx = 1 x sin 2x + 1 cos 2x + C.

2

2

2

4

12. x sin(3 - x) dx solution Let u = x and v = sin(3 - x). Then we have

u = x v = cos(3 - x) u = 1 v = sin(3 - x) Using Integration by Parts, we get

x sin(3 - x) dx = x cos(3 - x) - (1) cos(3 - x) dx = x cos(3 - x) + sin(3 - x) + C

13. x2 sin x dx solution Let u = x2 and v = sin x. Then we have

u = x2 v = - cos x u = 2x v = sin x Using Integration by Parts, we get x2 sin x dx = x2(- cos x) - 2x(- cos x) dx = -x2 cos x + 2 x cos x dx.

We must apply Integration by Parts again to evaluate x cos x dx. Taking u = x and v = cos x, we get

x cos x dx = x sin x - sin x dx = x sin x + cos x + C. Plugging this into the original equation gives us

x2 sin x dx = -x2 cos x + 2(x sin x + cos x) + C = -x2 cos x + 2x sin x + 2 cos x + C.

14. x2 cos 3x dx

solution Let u = x2 and v = cos 3x. Then we have

u = x2 u = 2x

v = 1 sin 3x 3

v = cos 3x

Using Integration by Parts, we get

x2 cos 3x dx = 1 x2 sin 3x - (2x) 1 sin 3x dx = 1 x2 sin 3x - 2

3

3

3

3

Use Integration by Parts again on this integral, with u = x and v = sin 3x to get

x sin 3x dx

x2 cos 3x dx = 1 x2 sin 3x - 2 - 1 x cos 3x + 1 cos 3x dx

3

33

3

= 1 x2 sin 3x + 2 x cos 3x - 2 sin 3x + C

3

9

27

March 30, 2011

S E C T I O N 7.1 Integration by Parts 811

15. e-x sin x dx solution Let u = e-x and v = sin x. Then we have

u = e-x v = - cos x u = -e-x v = sin x Using Integration by Parts, we get e-x sin x dx = -e-x cos x - (-e-x )(- cos x) dx = -e-x cos x - e-x cos x dx.

We must apply Integration by Parts again to evaluate e-x cos x dx. Using u = e-x and v = cos x, we get

e-x cos x dx = e-x sin x - (-e-x )(sin x) dx = e-x sin x + e-x sin x dx. Plugging this into the original equation, we get

e-x sin x dx = -e-x cos x - e-x sin x + e-x sin x dx .

Solving this equation for e-x sin x dx gives us

e-x sin x dx = - 1 e-x (sin x + cos x) + C. 2

16. ex sin 2x dx solution Let u = sin 2x and v = ex . Then we have

u = sin 2x v = ex u = 2 cos 2x v = ex Using Integration by Parts, we get ex sin 2x dx = ex sin 2x - 2 ex cos 2x dx.

We must apply Integration by Parts again to evaluate ex cos 2x dx. Using u = cos 2x and v = ex , we get

ex cos 2x dx = ex cos 2x - (-2 sin 2x)ex dx = ex cos 2x + 2 ex sin 2x dx. Plugging this into the original equation, we get

ex sin 2x dx = ex sin 2x - 2 ex cos 2x + 2 ex sin 2x dx = ex sin 2x - 2ex cos 2x - 4 ex sin 2x dx.

Solving this equation for ex sin 2x dx gives us

ex sin 2x dx = 1 ex (sin 2x - 2 cos 2x) + C. 5

17. e-5x sin x dx

solution Let u = sin x and v = e-5x . Then we have

u = sin x u = cos x

v = - 1 e-5x 5

v = e-5x

Using Integration by Parts, we get

e-5x sin x dx = - 1 e-5x sin x - cos x - 1 e-5x dx = - 1 e-5x sin x + 1 e-5x cos x dx

5

5

5

5

March 30, 2011

812 C H A P T E R 7

TECHNIQUES OF INTEGRATION

Apply Integration by Parts again to this integral, with u = cos x and v = e-5x to get

e-5x cos x dx = - 1 e-5x cos x - 1 e-5x sin x dx

5

5

Plugging this into the original equation, we get

e-5x sin x dx = - 1 e-5x sin x + 1 - 1 e-5x cos x - 1 e-5x sin x dx

5

55

5

= - 1 e-5x sin x - 1 e-5x cos x - 1 e-5x sin x dx

5

25

25

Solving this equation for e-5x sin x dx gives us

e-5x sin x dx = - 5 e-5x sin x - 1 e-5x cos x + C = - 1 e-5x (5 sin x + cos x) + C

26

26

26

18. e3x cos 4x dx

solution Let u = cos 4x and v = e3x . Then we have

u = cos 4x u = -4 sin 4x

v = 1 e3x 3

v = e3x

Using Integration by Parts, we get

e3x cos 4x dx = 1 e3x cos 4x - 1 e3x (-4 sin 4x) dx = 1 e3x cos 4x + 4

3

3

3

3

Apply Integration by Parts again to this integral, with u = sin 4x and v = e3x , to get

e3x sin 4x dx

e3x sin 4x dx = 1 e3x sin 4x - 1 e3x ? 4 cos 4x dx = 1 e3x sin 4x - 4 e3x cos 4x dx

3

3

3

3

Plugging this into the original equation, we get

e3x cos 4x dx = 1 e3x cos 4x + 4 1 e3x sin 4x - 4 e3x cos 4x dx

3

33

3

= 1 e3x cos 4x + 4 e3x sin 4x - 16 e3x cos 4x dx

3

9

9

Solving this equation for e3x cos 4x dx gives us

e3x cos 4x dx = 3 e3x cos 4x + 4 e3x sin 4x = 1 e3x (3 cos 4x + 4 sin 4x) + C

25

25

25

19. x ln x dx

solution Let u = ln x and v = x. Then we have

Using Integration by Parts, we get

u = ln x

v

=

1 2

x2

u

=

1 x

v =x

x ln x dx = 1 x2 ln x - 2

= 1 x2 ln x - 1

2

2

1 1x2 dx x2

x dx = 1 x2 ln x - 1

2

2

x2 + C = 1 x2(2 ln x - 1) + C.

2

4

March 30, 2011

S E C T I O N 7.1 Integration by Parts 813

20.

ln x dx x2

solution Let u = ln x and v = x-2. Then we have

u = ln x v = -x-1

u

=

1 x

v = x-2

Using Integration by Parts, we get

ln x x2

dx

=

-

1 x

ln

x

-

1 x

-1 x

dx

=

-

1 x

ln

x

+

=

-

1 x

ln

x

-

1 x

+

C

=

-

1 x

(ln

x

+

1)

+

C.

x-2 dx

21. x2 ln x dx

solution Let u = ln x and v = x2. Then we have

Using Integration by Parts, we get

u = ln x

v

=

1 3

x3

u

=

1 x

v = x2

x2 ln x dx = 1 x3 ln x - 3

1 x

1 x3 3

dx = 1 x3 ln x - 1

3

3

x2 dx

= 1 x3 ln x - 1 x3 + C = x3 ln x - 1 + C.

3

33

3

3

22. x-5 ln x dx

solution Let u = ln x and v = x-5. Then we have

Using Integration by Parts, we get

u = ln x

u

=

1 x

v = - 1 x-4 4

v = x-5

x-5 ln x dx = - 1 x-4 ln x + 4

1 x-4 4

1 x

dx

=

- 1 x-4 4

ln x

+

1 4

x-5 dx

=

-

1 4

x-4

ln

x

-

1 x-4 16

+

C

=

-

1 4x4

ln x + 1 4

+C

23. (ln x)2 dx

solution Let u = (ln x)2 and v = 1. Then we have

Using Integration by Parts, we get

u = (ln x)2 v = x

u

=

2 x

ln x

v =1

(ln x)2 dx = (ln x)2(x) -

2 x

ln x

x dx = x(ln x)2 - 2

ln x dx.

We must apply Integration by Parts again to evaluate ln x dx. Using u = ln x and v = 1, we have

ln x dx = x ln x - Plugging this into the original equation, we get

1 x

?

x

dx

=

x

ln

x

-

dx = x ln x - x + C.

(ln x)2 dx = x(ln x)2 - 2 (x ln x - x) + C = x (ln x)2 - 2 ln x + 2 + C.

March 30, 2011

814 C H A P T E R 7 TECHNIQUES OF INTEGRATION

24. x(ln x)2 dx

solution Let u = (ln x)2, v = x. Then we have

u = (ln x)2

u

=

2 ln x x

v = 1x2 2

v =x

Using Integration by Parts, we get

x(ln x)2 dx = 1 x2(ln x)2 - 2

x2

ln x x

dx

=

1 x2(ln x)2 2

-

Apply Integration by Parts again to this integral, with u = ln x, v = x, to get

x ln x dx

x ln x dx = 1 x2 ln x - 1

2

2

Plug this back into the first formula to get

x2

1 x

dx

=

1 x2 2

ln

x

-

1 x2 4

x(ln x)2 dx = 1 x2(ln x)2 - 1 x2 ln x - 1 x2 + C = 1 x2 (ln x)2 - ln x + 1 + C

2

2

4

2

2

25. x sec2 x dx solution Let u = x and v = sec2 x. Then we have

u = x v = tan x u = 1 v = sec2 x Using Integration by Parts, we get

x sec2 x dx = x tan x - (1) tan x dx = x tan x - ln | sec x| + C.

26. x tan x sec x dx solution Let u = x and v = tan x sec x. Then we have

u = x v = sec x u = 1 v = tan x sec x Using Integration by Parts, we get

x tan x sec x dx = x sec x - sec x dx = x sec x - ln |sec x + tan x| + C

27. cos-1 x dx

solution Let u = cos-1 x and v = 1. Then we have

u = cos-1 x u = -1

1 - x2

v=x v =1

Using Integration by Parts, we get

cos-1 x dx = x cos-1 x -

-x dx. 1 - x2

We can evaluate

-x dx by making the substitution w = 1 - x2. Then dw = -2x dx, and we have 1 - x2

cos-1 x dx = x cos-1 x - 1 2

-2x dx = x cos-1 x - 1 w-1/2 dw

1 - x2

2

= x cos-1 x - 1 (2w1/2) + C = x cos-1 x - 1 - x2 + C. 2

March 30, 2011

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download