Download TRIGONOMETRIC INTEGRALS In this section we use trigonometric

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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 y cos 3x dx.
Simply substituting u ­ cos x isn’t helpful, since then du ­ 2sin 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
sin 2x 1 cos 2x ­ 1:
cos 3x ­ cos 2x ? cos x ­ s1 2 sin 2xd cos x
SOLUTION
We can then evaluate the integral by substituting u ­ sin x, so du ­ cos x dx and
y cos 3x dx ­ y cos 2x ? cos x dx ­ y s1 2 sin 2xd cos x dx
­ y s1 2 u 2 d du ­
­ sin
u 2 13 u 3 1 C
x 2 13 sin 3x 1 C
M
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 sin 2x 1 cos 2x ­ 1 enables us to convert back and forth between even powers
of sine and cosine.
V EXAMPLE 2
Find y sin 5x cos 2x dx.
We could convert cos 2x to 1 2 sin 2x, 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 sin 4x factor in terms of cos x :
SOLUTION
sin 5x cos 2x ­ ssin2xd2 cos 2x sin x ­ s1 2 cos 2xd2 cos 2x sin x
N
Figure 1 shows the graphs of the integrand
sin 5x cos 2x in Example 2 and its indefinite integral (with C ­ 0). Which is which?
Substituting u ­ cos x, we have du ­ 2sin x dx and so
y sin 5x cos 2x dx ­ y ssin 2xd 2 cos 2x sin x dx
0.2
­ y s1 2
_π
π
_0.2
FIGURE 1
cos 2xd2 cos 2x sin x dx
­ y s1 2 u 2 d2 u 2 s2dud ­ 2y su 2 2
S3
­2
u3
2
2
u5
1
u7
D
1
2u 4 1 u 6 d du
C
5
7
1
2
1
3
5
­ 2 3 cos x 1 5 cos x 2 7 cos 7x 1 C
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SECTION 7.2 TRIGONOMETRIC INTEGRALS
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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):
sin 2x ­ 12 s1 2 cos 2 xd
cos 2x ­ 12 s1 1 cos 2 xd
and
Evaluate y0 sin 2x dx.
p
Example 3 shows that the area of the region
shown in Figure 2 is py 2.
N
V EXAMPLE 3
If we write sin 2x ­ 1 2 cos 2x, the integral is no simpler to evaluate. Using the
half-angle formula for sin 2x, however, we have
SOLUTION
1.5
y=sin@€x
0
_0.5
FIGURE 2
p
p
p
y0 sin 2x dx ­ 12 y0 s1 2 cos 2 xd dx ­ [ 12 ( x 2 12 sin 2 x)]0
1
(
)
1
1
­ 2 p 2 2 sin 2p 2 2
π
(0 2 12 sin 0) ­ 12 p
Notice that we mentally made the substitution u ­ 2 x when integrating cos 2 x. Another
M
method for evaluating this integral was given in Exercise 43 in Section 7.1.
EXAMPLE 4
Find y sin 4x dx.
We could evaluate this integral using the reduction formula for x sin n x dx
(Equation 7.1.7) together with Example 3 (as in Exercise 43 in Section 7.1), but a better
method is to write sin 4x ­ ssin 2xd2 and use a half-angle formula:
SOLUTION
y sin 4x dx ­ y ssin 2xd2 dx
­
y
S1
2
cos 2 x
2
D
2
dx
­ 4 y s1 2 2 cos 2 x 1
1
cos 2 2 xd dx
Since cos 2 2 x occurs, we must use another half-angle formula
cos 2 2 x ­ 12 s1 1 cos 4 xd
This gives
y sin 4x dx ­ 14 y f1 2 2 cos 2 x 1 12 s1 1 cos 4 xdg dx
­4y
1
(
( 32 2 2 cos 2x 1 12 cos 4x) dx
1 3
2x
­4
1
)
2 sin 2 x 1 8 sin 4 x 1
C
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To summarize, we list guidelines to follow when evaluating integrals of the form
x sin mx cos nx dx, where m ù 0 and n ù 0 are integers.
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CHAPTER 7 TECHNIQUES OF INTEGRATION
y sin m x cos n x dx
(a) If the power of cosine is odd sn ­ 2 k 1 1d, save one cosine factor and use
cos 2x ­ 1 2 sin 2x to express the remaining factors in terms of sine:
STRATEGY FOR EVALUATING
y sin m x cos 2 k11x dx ­ y sin m x scos 2xdk cos x dx
­ y sin m x s1 2
sin 2xdk cos x dx
Then substitute u ­ sin x.
(b) If the power of sine is odd sm ­ 2 k 1 1d, save one sine factor and use
sin 2x ­ 1 2 cos 2x to express the remaining factors in terms of cosine:
y sin 2 k11x cos n x dx ­ y ssin 2xdk cos n x sin x dx
­ y s1 2 cos 2xdk cos n x
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
sin 2x ­ 12 s1 2 cos 2 xd
cos 2x ­ 12 s1 1 cos 2 xd
It is sometimes helpful to use the identity
sin x cos x ­ 12 sin 2 x
We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
sdydxd tan x ­ sec 2x, we can separate a sec 2x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x ­ 1 1 tan 2x.
Or, since sdydxd 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 y tan 6x sec 4x dx.
SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in
terms of tangent using the identity sec 2x ­ 1 1 tan 2x. We can then evaluate the integral
by substituting u ­ tan x so that du ­ sec 2x dx :
y tan 6x sec 4x dx ­ y tan 6x sec 2x sec 2 x dx
­ y tan 6x s1 1
­ y u 6s1 1
u7
1
u9
tan 2xd sec 2x dx
u 2 d du ­ y su 6 1 u 8 d du
1C
7
9
1
1
­ 7 tan 7x 1 9 tan 9x 1 C
­
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SECTION 7.2 TRIGONOMETRIC INTEGRALS
EXAMPLE 6
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463
Find y tan 5u sec 7u du.
If we separate a sec 2u factor, as in the preceding example, we are left with
a sec u factor, which isn’t easily converted to tangent. However, if we separate a
sec u tan u factor, we can convert the remaining power of tangent to an expression
involving only secant using the identity tan 2u ­ sec 2u 2 1. We can then evaluate the
integral by substituting u ­ sec u, so du ­ sec u tan u du :
SOLUTION
5
y tan 5u sec 7u du ­ y tan 4u sec 6u sec u tan u du
­ y ssec 2u 2 1d2 sec 6u
sec u tan u du
­ y su 2 2 1d2 u 6 du
­ y su 10 2
u 11
22
2u 8 1 u 6 d du
u9
1
u7
1C
11
9
7
1
2
1
­ 11 sec 11u 2 9 sec 9u 1 7 sec 7u 1 C
­
M
The preceding examples demonstrate strategies for evaluating integrals of the form
x tan mx sec nx dx for two cases, which we summarize here.
y tan mx sec nx dx
(a) If the power of secant is even sn ­ 2 k, k ù 2d, save a factor of sec 2x and use
sec 2x ­ 1 1 tan 2x to express the remaining factors in terms of tan x :
STRATEGY FOR EVALUATING
y tan m x sec 2 kx dx ­ y tan m x ssec 2xdk21 sec 2x dx
­y
tan m x s1 1 tan 2xdk21 sec 2x dx
Then substitute u ­ tan x.
(b) If the power of tangent is odd sm ­ 2 k 1 1d, save a factor of sec x tan x and
use tan 2x ­ sec 2x 2 1 to express the remaining factors in terms of sec x :
y tan 2 k11x sec n x dx ­ y stan 2xdk sec n21x sec x tan x dx
­ y ssec 2x 2 1dk sec n21x
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
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CHAPTER 7 TECHNIQUES OF INTEGRATION
integrate tan x by using the formula established in (5.5.5):
y tan x dx ­ ln | sec x | 1 C
We will also need the indefinite integral of secant:
y sec x dx ­ ln | sec x 1 tan x | 1 C
1
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x 1 tan x :
sec x 1 tan x
y sec x dx ­ y sec x sec x 1 tan x dx
­y
sec 2x 1 sec x tan x
dx
sec x 1 tan x
If we substitute u ­ sec x 1 tan x, then du ­ ssec x tan x 1 sec 2xd dx, so the integral
becomes x s1yud du ­ ln | u | 1 C. Thus we have
y sec x dx ­ ln | sec x 1 tan x | 1 C
EXAMPLE 7
Find y tan 3x dx.
Here only tan x occurs, so we use tan 2x ­ sec 2x 2 1 to rewrite a tan 2x factor in
terms of sec 2x :
SOLUTION
y tan 3x dx ­ y tan x tan 2x dx ­ y tan x ssec 2x 2 1d dx
­y
tan x sec 2x dx 2 y tan x dx
tan 2x
2 ln | sec x | 1 C
2
In the first integral we mentally substituted u ­ tan x so that du ­ sec 2x dx.
­
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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
SOLUTION
Find y sec 3x dx.
Here we integrate by parts with
u ­ sec x
du ­ sec x tan x dx
dv ­ sec 2x dx
v ­ tan x
SECTION 7.2 TRIGONOMETRIC INTEGRALS
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465
y sec 3x dx ­ sec x tan x 2 y sec x tan 2x dx
Then
­
sec x tan x 2 y sec x ssec 2x 2 1d dx
­ sec
x tan x 2 y sec 3x dx 1 y sec x dx
Using Formula 1 and solving for the required integral, we get
y sec 3x dx ­ 12 (sec x tan x 1 ln | sec x 1 tan x |) 1 C
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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 x cot m x csc n x dx can be found by similar methods because of the identity
1 1 cot 2x ­ csc 2x.
Finally, we can make use of another set of trigonometric identities:
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 ­ 12 fsinsA 2 Bd 1 sinsA 1 Bdg
These product identities are discussed in
Appendix D.
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(b) sin A sin B ­ 12 fcossA 2 Bd 2 cossA 1 Bdg
(c) cos A cos B ­ 12 fcossA 2 Bd 1 cossA 1 Bdg
EXAMPLE 9
Evaluate y sin 4 x 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 4 x cos 5x dx ­ y 21 fsins2xd 1 sin 9xg dx
­ 2 y s2sin x 1
1
sin 9xd dx
1
(cos x 2 19 cos 9xd 1 C
­2
7.2
1– 49
1.
EXERCISES
Evaluate the integral.
y sin x cos x dx
3
3py4
2
2.
y sin x cos x dx
6
py2
3.
ypy2 sin x cos x dx
4.
y0 cos x dx
5.
y sin 2 sp xd cos 5 sp xd dx
6.
y
7.
py2
y0 cos2u du
8.
py2
y0 sin 2 s2u d du
5
3
p
y0 sin 4s3 td dt
10.
p
y0 cos6u du
11.
y s1 1 cos u d2 du
12.
y x cos2x dx
13.
py2
y0 sin 2x cos 2x dx
14.
p
y0 sin 2 t cos 4 t dt
15.
cos a
y ssin a da
16.
y cos u cos5ssin u d du
9.
3
5
sin3 (sx )
sx
dx
5
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CHAPTER 7 TECHNIQUES OF INTEGRATION
17.
y cos 2x tan 3x dx
19.
y
21.
cos x 1 sin 2 x
dx
sin x
y sec 2 x tan x dx
18.
y cot 5u sin 4u du
20.
y cos 2 x sin 2 x dx
22.
py 2
y0 sec 4s ty 2d dt
23.
y tan 2x dx
24.
y stan 2 x 1 tan 4 xd dx
25.
y sec 6 t dt
26.
py4
y0 sec 4u tan 4u du
27.
py3
y0 tan 5 x sec 4 x dx
28.
y tan 3s 2 xd sec 5s 2 xd dx
29.
y tan3x sec x dx
30.
py3
y0 tan 5x sec 6 x dx
31.
y tan 5x dx
33.
53.
55.
56.
y ­ sin3x, y ­ cos 3 x, py4 ø x ø 5py4
y tan 2x sec x dx
35.
y x sec x tan x dx
36.
y cos3 f df
37.
ypy6 cot 2x dx
38.
ypy4 cot 3x dx
39.
y cot 3a csc 3a da
40.
y csc 4 x cot 6 x dx
41.
y csc x dx
42.
ypy6 csc 3x dx
43.
y sin 8 x cos 5x dx
44.
y cos p x cos 4p x dx
45.
y sin 5u sin u du
46.
y
cos x 1 sin x
dx
sin 2 x
65.
47.
1 2 tan x
y sec 2x dx
48.
49.
y t sec 2s t 2 d tan 4s t 2 d dt
2
50.
py3
59.
52.
y sin 3x cos 4 x dx
2p
y0 cos 3x dx
60.
2
y0 sin 2p x cos 5p x dx
Find the volume obtained by rotating the region bounded
by the given curves about the specified axis.
61. y ­ sin x, y ­ 0, py 2 ø x ø p ; about the x-axis
61–64
62.
y ­ sin 2 x, y ­ 0, 0 ø x ø p ; about the x-axis
63.
y ­ sin x, y ­ cos x, 0 ø x ø py4; about y ­ 1
64.
y ­ sec x, y ­ cos x, 0 ø x ø py3; about y ­ 21
66.
your answer is reasonable, by graphing both the integrand and its
antiderivative (taking C ­ 0d.
y x sin 2 s x 2 d dx
integral. Then use the methods of this section to prove that your
guess is correct.
dx
; 51–54 Evaluate the indefinite integral. Illustrate, and check that
51.
; 59–60 Use a graph of the integrand to guess the value of the
y cos x 2 1
If x0py 4 tan 6 x sec x dx ­ I , express the value of
x0py 4 tan 8 x sec x dx in terms of I.
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 2 x ­ 2 sin x cos x
(d) integration by parts
Explain the different appearances of the answers.
58.
34.
py2
Find the average value of the function f s xd ­ sin 2x cos 3x on
the interval f2p, pg.
57.
y cos 4u du
py2
x
y sec 4 2 dx
Find the area of the region bounded by the given curves.
y ­ sin 2 x, y ­ cos 2 x, 2py4 ø x ø py4
y tan 6sayd dy
sin f
54.
57–58
32.
tan 3u
y sin 3x sin 6 x dx
A particle moves on a straight line with velocity function
vstd ­ sin v t cos 2v t. Find its position function s ­ f std
if f s0d ­ 0.
Household electricity is supplied in the form of alternating
current that varies from 155 V to 2155 V with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
Estd ­ 155 sins120p td
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 fEstdg 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 Estd ­ A sins120p td.