Calculus 2nd Edition Rogawski

SECTION 8.2 Trigonometric Integrals 419
Integrating sin m x cos n x
Case 1: m = 2k + 1 odd
Write sin 2k+1 x as (1 − cos 2 x) k sin x.
Then ∫ sin 2k+1 x cos n xdxbecomes
∫
sin x(1 − cos 2 x) k cos n xdx
Substitute u = cos x,−du = sin xdx.
Case 2: n = 2k + 1 odd
Write cos 2k+1 x as (1 − sin 2 x) k cos x
Then ∫ sin m x cos 2k+1 xdxbecomes
∫
sin m x(1 − sin 2 x) k cos xdx
Substitute u = sin x,du = cos xdx.
Case 3: m, n both even
Use reduction formulas (1) or (2) as
described below or use the method of
Exercises 65–68.
The strategy of the previous example works when sin m x appears with m odd. Similarly,
if n is odd, write cos n x as a power of (1 − sin 2 x) times cos x.
EXAMPLE 2 Odd Power of sin x or cos x
∫
Evaluate
sin 4 x cos 5 xdx.
Solution We take advantage of the fact that cos 5 x is an odd power to write
sin 4 x cos 5 xdx= sin 4 x cos 4 x(cos xdx)= sin 4 x(1 − sin 2 x) 2 (cos xdx)
This allows us to use the substitution u = sin x, du = cos xdx:
∫
∫
sin 4 x cos 5 xdx= (sin 4 x)(1 − sin 2 x) 2 cos xdx
∫
∫
= u 4 (1 − u 2 ) 2 du = (u 4 − 2u 6 + u 8 )du
= u5
5 − 2u7
7 + u9
9 + C = sin5 x
− 2 sin7 x
+ sin9 x
+ C
5 7 9
The following reduction formulas can be used to integrate sin n x and cos n x for any
exponent n, even or odd (their proofs are left as exercises; see Exercise 64).
Reduction Formulas for Sine and Cosine
∫
sin n xdx= − 1 n sinn−1 x cos x + n − 1
n
∫
sin n−2 xdx 1
∫
cos n xdx= 1 n cosn−1 x sin x + n − 1
n
∫
cos n−2 xdx 2
∫
EXAMPLE 3 Evaluate
sin 4 xdx.
Solution Apply Eq. (1) with n = 4,
∫
sin 4 xdx= − 1 4 sin3 x cos x + 3 ∫
4
sin 2 xdx 3
Then apply Eq. (1) again, with n = 2, to the integral on the right:
∫
sin 2 xdx= − 1 2 sin x cos x + 1 ∫
dx = − 1 2 2 sin x cos x + 1 2 x + C 4
Using Eq. (4) in Eq. (3), we obtain
∫
sin 4 xdx= − 1 4 sin3 x cos x − 3 8 sin x cos x + 3 8 x + C
Trigonometric integrals can be expressed in many different ways because trigonometric
functions satisfy a large number of identities. For example, a computer algebra
system might evaluate the integral in the previous example as
∫
sin 4 xdx= 1 (x − 8 sin 2x + sin 4x) + C
32
You can check that this agrees with the result in Example 3 (Exercise 61).