Trigonometric Integrals

[Pages:12]Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.

y EXAMPLE 1 Evaluate cos3x dx.

SOLUTION Simply substituting u cos x isn't helpful, since then du sin x dx. In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor. Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin2x cos2x 1:

cos3x cos2x cos x 1 sin2x cos x

We can then evaluate the integral by substituting u sin x, so du cos x dx and

y cos3x dx y cos2x cos x dx y 1 sin2x cos x dx

y

1

u2

du

u

1 3

u3

C

sin

x

1 3

sin3x

C

In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity sin2x cos2x 1 enables us to convert back and forth between even powers of sine and cosine.

y EXAMPLE 2 Find sin5x cos2x dx

SOLUTION We could convert cos2x to 1 sin2x, but we would be left with an expression in terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x :

sin5x cos2x sin2x2 cos2x sin x 1 cos2x2 cos2x sin x

Figure 1 shows the graphs of the integrand sin5x cos2x in Example 2 and its indefinite integral (with C 0). Which is which?

0.2

Substituting u cos x, we have du sin x dx and so

y y sin5x cos2x dx sin2x2 cos2x sin x dx y 1 cos2x2 cos2x sin x dx

_

_ 0.2 FIGURE 1

y y 1 u 2 2 u 2 du u 2 2u 4 u 6 du

u3

u5 u7

2 C

3

57

1 3

cos3x

2 5

cos5x

1 7

cos7x

C

1

2 TRIGONOMETRIC INTEGRALS

In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix C):

sin2x

1 2

1

cos

2x

and

cos2x

1 2

1

cos

2x

Example 3 shows that the area of the region shown in Figure 2 is 2.

y EXAMPLE 3 Evaluate sin2x dx. 0

SOLUTION If we write sin2x 1 cos2x, the integral is no simpler to evaluate. Using the half-angle formula for sin2x, however, we have

y y [ ] ( )

sin2x

dx

1 2

1 cos 2x dx

1 2

x

1 2

sin

2x

0

0

0

( ) ( )

1 2

1 2

sin

2

1 2

0

1 2

sin

0

1 2

Notice that we mentally made the substitution u 2x when integrating cos 2x. Another method for evaluating this integral was given in Exercise 33 in Section 5.6.

1.5 y=sin@ x

0

FIGURE 2

_ 0.5

y EXAMPLE 4 Find sin4x dx.

SOLUTION We could evaluate this integral using the reduction formula for x sinnx dx

(Equation 5.6.7) together with Example 3 (as in Exercise 33 in Section 5.6), but a better method is to write sin4x sin2x2 and use a half-angle formula:

y y sin4x dx sin2x2 dx

1 cos 2x 2

y 2

dx

y

1 4

1 2 cos 2x cos2 2x dx

Since cos2 2x occurs, we must use another half-angle formula

This gives

cos2

2x

1 2

1

cos

4x

y y sin4x

dx

1 4

1

2

cos

2x

1 2

1

cos

4x

dx

y

1 4

(3 2

2

cos

2x

1 2

cos

4x)

dx

1 4

(3 2

x

sin

2x

1 8

sin

4x)

C

To summarize, we list guidelines to follow when evaluating integrals of the form

x sinmx cosnx dx, where m 0 and n 0 are integers.

TRIGONOMETRIC INTEGRALS 3

y Strategy for Evaluating sinmx cosnx dx

(a) If the power of cosine is odd n 2k 1, save one cosine factor and use cos2x 1 sin2x to express the remaining factors in terms of sine:

y y sinmx cos2k1x dx sinmx cos2xk cos x dx

y sinmx 1 sin2xk cos x dx

Then substitute u sin x. (b) If the power of sine is odd m 2k 1, save one sine factor and use

sin2x 1 cos2x to express the remaining factors in terms of cosine:

y y sin2k1x cosnx dx sin2xk cosnx sin x dx

y 1 cos2xk cosnx sin x dx

Then substitute u cos x. [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.]

(c) If the powers of both sine and cosine are even, use the half-angle identities

sin2x

1 2

1

cos

2x

cos2x

1 2

1

cos

2x

It is sometimes helpful to use the identity

sin

x

cos

x

1 2

sin

2x

We can use a similar strategy to evaluate integrals of the form x tanmx secnx dx. Since

ddx tan x sec2x, we can separate a sec2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x 1 tan2x. Or, since ddx sec x sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant.

y EXAMPLE 5 Evaluate tan6x sec4x dx.

SOLUTION If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x 1 tan2x. We can then evaluate the integral by substituting u tan x with du sec2x dx:

y y tan6x sec4x dx tan6x sec2x sec2x dx

y tan6x 1 tan2x sec2x dx

y y u 61 u 2 du u 6 u 8 du

u7 u9 C 79

1 7

tan7x

1 9

tan9x

C

4 TRIGONOMETRIC INTEGRALS

y EXAMPLE 6 Find tan5 sec7 d.

SOLUTION If we separate a sec2 factor, as in the preceding example, we are left with a sec5 factor, which isn't easily converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression involving only secant using the identity tan2 sec2 1. We can then evaluate the integral by substituting u sec , so du sec tan d :

y y tan5 sec7 d tan4 sec6 sec tan d

y sec2 12 sec6 sec tan d

y y u 2 12 u 6 du u 10 2u 8 u 6 du

u 11

u9 u7

2 C

11

97

1 11

sec11

2 9

sec9

1 7

sec7

C

The preceding examples demonstrate strategies for evaluating integrals of the form

x tanmx secnx dx for two cases, which we summarize here.

y Strategy for Evaluating tanmx secnx dx

(a) If the power of secant is even n 2k, k 2, save a factor of sec2x and use sec2x 1 tan2x to express the remaining factors in terms of tan x :

y y tanmx sec2kx dx tanmx sec2xk1 sec2x dx

y tanmx 1 tan2xk1 sec2x dx

Then substitute u tan x. (b) If the power of tangent is odd m 2k 1, save a factor of sec x tan x and

use tan2x sec2x 1 to express the remaining factors in terms of sec x :

y y tan2k1x secnx dx tan2xk secn1x sec x tan x dx

y sec2x 1k secn1x sec x tan x dx

Then substitute u sec x.

For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to integrate tan x by using the formula established in Example 5 in Section 5.5:

y tan x dx ln sec x C

We will also need the indefinite integral of secant:

TRIGONOMETRIC INTEGRALS 5

1

y sec x dx ln sec x tan x C

We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :

y

sec

x

dx

y

sec

x

sec sec

x x

tan tan

x x

dx

y

sec2x sec x tan sec x tan x

x

dx

If we substitute u sec x tan x, then du sec x tan x sec2x dx, so the integral

becomes x 1u du ln u C. Thus, we have

y sec x dx ln sec x tan x C

y EXAMPLE 7 Find tan3x dx.

SOLUTION Here only tan x occurs, so we use tan2x sec2x 1 to rewrite a tan2x factor in terms of sec2x :

y tan3x dx y tan x tan2x dx

y tan x sec2x 1 dx

y tan x sec2x dx y tan x dx

tan2x

ln sec x C

2

In the first integral we mentally substituted u tan x so that du sec2x dx.

If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. Powers of sec x may require integration by parts, as shown in the following example.

y EXAMPLE 8 Find sec3x dx.

SOLUTION Here we integrate by parts with u sec x du sec x tan x dx

dv sec2x dx v tan x

Then

y sec3x dx sec x tan x y sec x tan2x dx

sec x tan x y sec x sec2x 1 dx

sec x tan x y sec3x dx y sec x dx

6 TRIGONOMETRIC INTEGRALS

Using Formula 1 and solving for the required integral, we get

y sec3x

dx

1 2

(sec

x

tan

x

ln

sec x tan x ) C

Integrals such as the one in the preceding example may seem very special but they

occur frequently in applications of integration, as we will see in Chapter 6. Integrals of

the form x cotmx cscnx dx can be found by similar methods because of the identity

1 cot2x csc2x. Finally, we can make use of another set of trigonometric identities:

These product identities are discussed in Appendix C.

2 To evaluate the integrals (a) x sin mx cos nx dx, (b) x sin mx sin nx dx, or (c) x cos mx cos nx dx, use the corresponding identity:

(a)

sin

A

cos

B

1 2

sinA

B

sinA

B

(b)

sin

A

sin

B

1 2

cosA

B

cosA

B

(c)

cos A

cos B

1 2

cosA

B

cosA

B

y EXAMPLE 9 Evaluate sin 4x cos 5x dx.

SOLUTION This integral could be evaluated using integration by parts, but it's easier to use the identity in Equation 2(a) as follows:

y

sin

4x

cos

5x

dx

y

1 2

sinx

sin

9x

dx

y

1 2

sin x sin 9x dx

1 2

(cos

x

1 9

cos

9x

C

Exercises

A Click here for answers. 1?47 Evaluate the integral.

y 1. sin3x cos2x dx y 3. 34 sin5x cos3x dx

2

y 5. cos5x sin4x dx y 7. 2 cos2 d

0

y 9. sin43t dt 0

y 11. 1 cos 2 d y 13. 4 sin4x cos2x dx

0

S Click here for solutions.

y 2. sin6x cos3x dx y 4. 2 cos5x dx

0

y 6. sin3mx dx y 8. 2 sin22 d

0

y 10. cos6 d 0

y 12. x cos2x dx y 14. 2 sin2x cos2x dx

0

y 15. sin3x scos x dx

y 17. cos2x tan3x dx

19.

y

1 sin x cos x

dx

y 21. sec2x tan x dx

y 23. tan2x dx

y 25. sec6t dt y 27. 3 tan5x sec4x dx

0

y 29. tan3x sec x dx

y 31. tan5x dx

16. y cos cos5sin d y 18. cot5 sin4 d y 20. cos2x sin 2x dx y 22. 2 sec4t2 dt

0

y 24. tan4x dx y 26. 4 sec4 tan4 d

0

y 28. tan32x sec52x dx y 30. 3 tan5x sec6x dx

0

y 32. tan6ay dy

TRIGONOMETRIC INTEGRALS 7

y 33.

tan3 cos4

d

y 34. tan2x sec x dx

; 57?58 Use a graph of the integrand to guess the value of the integral. Then use the methods of this section to prove that your guess is correct.

y 35. 2 cot2x dx 6

y 36. 2 cot3x dx 4

y 57. 2 cos3x dx 0

y 58. 2 sin 2 x cos 5 x dx 0

y 37. cot 3 csc3 d 39. y csc x dx

y 38. csc 4x cot 6x dx y 40. 3 csc3x dx

6

59?62 Find the volume obtained by rotating the region bounded by the given curves about the specified axis. 59. y sin x, x 2, x , y 0; about the x-axis

60. y tan2x, y 0, x 0, x 4; about the x-axis

41. y sin 5x sin 2x dx

42. y sin 3x cos x dx

43. y cos 7 cos 5 d

1 tan2x

y 45.

sec2x dx

44.

y

cos x sin x sin 2x

dx

46.

y

dx cos x 1

y 47. t sec2t 2 tan4t 2 dt

48.

If

x 4 0

tan6x

sec x dx

I, express the value of

4

x 0

tan8x

sec

x

dx

in

terms

of

I.

; 49?52 Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking C 0.

y 49. sin5x dx

y 50. sin4x cos4x dx

51. y sin 3x sin 6x dx

y 52. sec4 x dx

2

53. Find the average value of the function f x sin2x cos3x on the interval , .

54. Evaluate x sin x cos x dx by four methods: (a) the substitution

u cos x, (b) the substitution u sin x, (c) the identity sin 2x 2 sin x cos x, and (d) integration by parts. Explain the different appearances of the answers.

55?56 Find the area of the region bounded by the given curves. 55. y sin x, y sin3x, x 0, x 2

56. y sin x, y 2 sin2x, x 0, x 2

61. y cos x, y 0, x 0, x 2; about y 1

62. y cos x, y 0, x 0, x 2; about y 1

63. A particle moves on a straight line with velocity function vt sin t cos2t. Find its position function s f t if f 0 0.

64. Household electricity is supplied in the form of alternating current that varies from 155 V to 155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation Et 155 sin120 t

where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the average value of Et2 over one cycle. (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V.

Find the corresponding amplitude A needed for the voltage Et A sin120t.

65?67 Prove the formula, where m and n are positive integers.

y 65. sin mx cos nx dx 0

y 66.

sin mx sin nx dx

0

if m n if m n

y 67.

cos mx cos nx dx

0

if m n if m n

68. A finite Fourier series is given by the sum

N

f x an sin nx n1

a1 sin x a2 sin 2x aN sin Nx

Show that the mth coefficient am is given by the formula

y 1

am

f x sin mx dx

8 TRIGONOMETRIC INTEGRALS

Answers

S Click here for solutions.

1.

1 5

cos5x

1 3

cos3x

C

3.

11 384

5.

1 5

sin5x

2 7

sin7x

1 9

sin9x

C

7. 4

9. 38

11.

3 2

2

sin

1 4

sin

2

C

13. 3 4192

15.

(2 7

cos3x

2 3

cos

x)

scos

x

C

17.

1 2

cos2x

ln

cos x

C

19. ln1 sin x C

21.

1 2

tan2x

C

23. tan x x C

25.

1 5

tan5t

2 3

tan3t

tan

t

C

27.

117 8

29.

1 3

sec3x

sec x

C

31.

1 4

sec4x

tan2x

ln

sec x

C

33.

1 6

tan6

1 4

tan4

C

35. s3 3

37.

1 3

csc3

1 5

csc5

C

39. ln csc x cot x C

41.

1 6

sin

3x

1 14

sin

7x

C

43.

1 4

sin 2

1 24

sin

12

C

45.

1 2

sin 2x

C

47.

1 10

tan5t 2

C

49.

1 5

cos5x

2 3

cos3x

cos x

C

1.1

_2

F 2

1.1

51.

1 6

sin

3x

1 18

sin

9x

C

1

F

_2

2

53. 0

55.

1 3

57. 0

63. s 1 cos3t3

1

59. 24

61. 2 24

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