James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

480 Chapter 7 Techniques of Integration
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 2 x − 1 2 s1 2 cos 2xd and cos2 x − 1 2 s1 1 cos 2xd
Example 3 shows that the area of
the region shown in Figure 2 is y2.
1.5
y=sin@ x
Example 3 Evaluate y
0 sin2 x dx.
SOLUTION If we write sin 2 x − 1 2 cos 2 x, the integral is no simpler to evaluate. Using
the half-angle formula for sin 2 x, however, we have
y
0 sin2 x dx − 1 2 y
s1 2 cos 2xd dx
0
− f 1 2 (x 2 1 2 sin 2x)g 0
0
_0.5
FIGURE 2
π
− 1 2 s 2 1 2 sin 2d 2 1 2 s0 2 1 2 sin 0d − 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 7.1.47.
n
Example 4 Find y sin 4 x dx.
SOLUTION We could evaluate this integral using the reduction formula for y sin n x dx
(Equation 7.1.7) together with Example 3 (as in Exercise 7.1.47), but a better method is
to write sin 4 x − ssin 2 xd 2 and use a half-angle formula:
y sin 4 x dx − y ssin 2 xd 2 dx
2
1 2 cos 2x
− y S D dx
2
− 1 4 y s1 2 2 cos 2x 1 cos 2 2xd dx
Since cos 2 2x occurs, we must use another half-angle formula
This gives
cos 2 2x − 1 2 s1 1 cos 4xd
y sin 4 x dx − 1 4 y f1 2 2 cos 2x 1 1 2 s1 1 cos 4xd g dx
− 1 4 y ( 3 2 2 2 cos 2x 1 1 2
cos 4x) dx
− 1 4 ( 3 2 x 2 sin 2x 1 1 8 sin 4x) 1 C
n
To summarize, we list guidelines to follow when evaluating integrals of the form
y sin m x cos n x dx, where m > 0 and n > 0 are integers.
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