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460 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

7.2 TRIGONOMETRIC INTEGRALS

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

EXAMPLE 1 Evaluate ycos3x 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
ycos3x dx ycos2x cos x dx y 1 sin2x cos x dx
y 1 u2 du u 31 u3 C
sin x 31 sin3x C	M

N 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

_π

_0.2

FIGURE 1

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.

V EXAMPLE 2 Find ysin5x 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 sin2x 2 cos2x sin x 1 cos2x 2 cos2x sin x

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

ysin5x cos2x dx y sin2x 2 cos2x sin x dx

y 1 cos2x 2 cos2x sin x dx

y 1 u2 2 u2 du y u2 2u4 u6 du

	3		5	7
		u3	u5	u7
			2		C
31 cos3x 52 cos5x 71 cos7x C						M

N Example 3 shows that the area of the region

shown in Figure 2 is		2.

1.5

y=sin@x

_0.5

FIGURE 2

SECTION 7.2 TRIGONOMETRIC INTEGRALS |||| 461

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 D):

	sin2x 21 1 cos 2x	and	cos2x 21 1 cos 2x

	EXAMPLE 3 Evaluate y0 sin2x dx.
V

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


y0 sin2x dx 21	y0	1 cos 2x dx [21 (x 21 sin 2x)]0
21	(	21 sin 2 ) 21 (0 21 sin 0) 21

Notice that we mentally made the substitution u 2x when integrating cos 2x. Another method for evaluating this integral was given in Exercise 43 in Section 7.1. M

EXAMPLE 4 Find ysin4x dx.

SOLUTION We could evaluate this integral using the reduction formula for xsinnx dx (Equation 7.1.7) together with Example 3 (as in Exercise 43 in Section 7.1), but a better method is to write sin4x sin2x 2 and use a half-angle formula:

ysin4x dx y sin2x 2 dx

y 1 cos2 2x 2 dx

14 y 1 2 cos 2x cos2 2x dx

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

cos2 2x 21 1		cos 4x
This gives
ysin4x dx 41	y 1 2 cos 2x 21 1 cos 4x dx
41	y(23 2 cos	2x 21 cos 4x) dx
41	(23 x sin 2x	81 sin 4x) C	M

To summarize, we list guidelines to follow when evaluating integrals of the form xsinmx cosnx dx, where m 0 and n 0 are integers.

462 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

STRATEGY FOR EVALUATING ysin mx cos nx 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:

ysinmx cos2k 1x dx ysinmx cos2x k cos x dx

ysinmx 1 sin2x k 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:

ysin2k 1x cosnx dx y sin2x k cosnx sin x dx

y 1 cos2x k 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.]

sin2x 21 1 cos 2x

cos2x 21 1 cos 2x

It is sometimes helpful to use the identity

sin x cos x 12 sin 2x

We can use a similar strategy to evaluate integrals of the form xtanmx secnx dx. Sinced dx 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 d dx 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.

V EXAMPLE 5 Evaluate ytan6x 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 so that du sec2x dx:

ytan6x sec4x dx ytan6x sec2x sec2x dx

ytan6x 1 tan2x sec2x dx

yu 6 1 u 2 du y u 6 u 8 du

u 7 u 9 C

7	9
71 tan7x 91 tan9x C		M

SECTION 7.2 TRIGONOMETRIC INTEGRALS |||| 463

EXAMPLE 6

Find ytan5

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 :

ytan5

sec7 d ytan4

sec6

sec tan d


y sec2					1 2 sec6 sec tan d
y u 2 1 2 u 6 du
y u 10 2u 8 u 6 du
		u 11				u 9	u 7
				2				C
	11				9		7

	1		sec11		92 sec9			71 sec7 C
	11

The preceding examples demonstrate strategies for evaluating integrals of the form xtanmx secnx dx for two cases, which we summarize here.

STRATEGY FOR EVALUATING ytanmx 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 :

ytanmx sec2kx dx ytanmx sec2x k 1 sec2x dx

ytanmx 1 tan2x k 1 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 :

ytan2k 1x secnx dx y tan2x k secn 1x sec x tan x dx

y sec2x 1 k secn 1x 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

464 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

integrate tan x by using the formula established in (5.5.5):

ytan x dx ln sec x C

We will also need the indeﬁnite integral of secant:

1	ysec 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 :

ysec x dx ysec x sec x tan x dx sec x tan x

y sec2x sec x tan x dx sec x tan x

If we substitute u sec x tan x, then du sec x tan x sec2x dx, so the integral becomes x 1 u du ln u C. Thus we have

ysec x dx ln sec x tan x C

EXAMPLE 7 Find ytan3x dx.

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

ytan3x dx ytan x tan2x dx ytan x sec2x 1 dx

ytan x sec2x dx ytan x dx

	tan	2x	ln sec x C
	2
In the ﬁrst integral we mentally substituted u tan x so that du sec2x dx.				M

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.

EXAMPLE 8 Find ysec3x dx.
SOLUTION Here we integrate by parts with
u sec x	dv sec2x dx
du sec x tan x dx	v tan x

	SECTION 7.2 TRIGONOMETRIC INTEGRALS \|\|\|\|	465
Then	ysec3x dx sec x tan x ysec x tan2x dx
	sec x tan x ysec x sec2x 1 dx
	sec x tan x ysec3x dx ysec x dx
Using Formula 1 and solving for the required integral, we get
	ysec3x dx 21 (sec x tan x ln sec x tan x ) C	M

N These product identities are discussed in Appendix D.

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 8. Integrals of the form xcotmx 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:

2 To evaluate the integrals (a) xsin mx cos nx dx, (b) xsin mx sin nx dx, or

(c)xcos mx cos nx dx, use the corresponding identity:

(a)sin A cos B 12 sin A B sin A B

(b)sin A sin B 12 cos A B cos A B

(c)cos A cos B 12 cos A B cos A B

EXAMPLE 9 Evaluate ysin 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:

7.2EXERCISES

1– 49 Evaluate the integral.

1.	ysin3x cos2x dx
	3 4
3.	y 2	sin5x cos3x dx
5.	ysin2 x cos5 x dx
	2
7.	y0	cos2 d

2.	ysin6x cos3x dx
	2
4.	y0		cos5x dx
			3

6.	y	sin (sx )			dx
		s
				x
	2
8.	y0		sin2 2 d

ysin 4x cos 5x dx y21 sin x sin 9x dx
21 y sin x sin 9x dx
21 (cos x 91 cos 9x C	M


9.	y0	sin4 3t dt			10.	y0	cos6 d
11.	y 1 cos 2 d				12.	yx cos2x dx
	2
13.	y0		sin2x cos2x dx		14.	y0	sin2 t cos4 t dt

15.	y	cos5		d	16.	ycos cos5 sin		d
		s
			sin

466 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

17.

x tan

x dx

18.

ycot

sin

ycos

19.

cos x sin 2x

20.

ycos2x sin 2x dx

sin x

ysec2x tan x dx

21.

22.

y0 2

sec4 t 2 dt

ytan2x dx

24.

y tan2 x tan4 x dx

23.

ysec6t dt

25.

26.

y0 4

sec4

tan4 d

ytan3 2x sec5 2x dx

27.

y0 3

tan5x sec4x dx

28.

29.

ytan3x sec x dx

30.

y0 3

tan5x sec6x dx

31.

ytan5x dx

32.

ytan6 ay dy

33.

tan3

34.

ytan2x sec x dx

cos4

35.

yx sec x tan x dx

36.

sin

cos3

37.

y 6

cot2x dx

38.

y 4

cot3x dx

39.

ycot 3 csc3 d

40.

ycsc 4x cot 6x dx

ycsc x dx

41.

42.

y 6

csc3x dx

ysin 8x cos 5x dx

44.

ycos

x cos 4 x dx

43.

45.

ysin 5

sin

46.

cos x sin x

sin 2x

47.

1 tan2x

48.

sec2x

cos x 1

49.

yt sec2 t 2 tan4 t 2 dt

50.

tan

x sec x dx I, express the value of

If x0

tan

x sec x dx in terms of I.

;51–54 Evaluate the indeﬁnite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking C 0 .

51. yx sin2 x 2 dx

52. ysin3x cos4 x dx

53. ysin 3x sin 6x dx	54. ysec4	x	dx
		2

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

56.Evaluate xsin 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

(d)integration by parts

Explain the different appearances of the answers.

57–58 Find the area of the region bounded by the given curves.

57.	y sin2 x,	y cos2 x,	4	x	4
58.	y sin3x,	y cos3 x,	4	x	5 4

;59–60 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.

2		2
59. y0 cos3x dx		60. y0 sin 2 x cos 5 x dx

61–64 Find the volume obtained by rotating the region bounded by the given curves about the speciﬁed axis.

		y sin x,	y 0,	2		x	; about the x-axis
	61.
62.		y sin2 x,	y 0,	0	x	;	about the x-axis
63.		y sin x,	y cos x,		0	x	4; about y 1
64.		y sec x,	y cos x,		0	x	3; about y 1

65.		A particle moves on a straight line with velocity function
		v t sin	t cos2 t. Find its position function s f t
		if f 0 0.

66.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

E t 155 sin 120 t

where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of E t 2 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 E t A sin 120 t .

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