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Trigonometric Techniques of Integration - (6.3)
1. Consider integrals of the form:
 sin n x cos m xdx
a. Either n or m is an odd positive integer:
n is an odd positive integer: n − 1 is even
 sin n x cos m xdx   sin n−1 x cos m x sin xdx  − 1 − cos 2 x n−1/2 cos m xdcos x
m is an odd positive integer: m − 1 is even
 sin n x cos m xdx   sin n x cos m−1 x cos xdx   sin n x1 − sin 2 x m−1/2 dsin x
b. Both n and m are positive even numbers:
Use the identities:
sin 2 x  1 1 − cos2x, and cos 2 x  1 1  cos2x
2
2
to reduce the power of sin x and cos x.
Example Evaluate the following integrals:
1.  cos 4 x sin 3 xdx
2.  sin 4 x cos 5 xdx
1  cos 4 x sin 3 x dx, u  cos x, du  − sin xdx
 cos 4 x sin 3 x dx  −  cos 4 x1 − cos 2 xdcos x  − u 4 − u 6 du
 − 1 u 5  1 u 7  C  − 1 cos 5 x  1 cos 7 x  C
7
5
7
5
4
4
2 2
5
2  sin x cos xdx   sin x1 − sin x dsin x, u  sin x, du  cos x
 sin 4 x cos 5 xdx   u 4 1 − 2u 2  u 4 du  u 4 − 2u 6  u 8 du
 1 u 5 − 2 u 7  1 u 9  C  1 sin 5 x − 2 sin 7 x  1 sin 9 x  C
7
5
7
5
9
9
Example Evaluate the following.
1
 cos 2 x sin 2 xdx
2
 cos 2 2x sin 4 2xdx
1
 cos 2 x sin 2 xdx  
1 1  cos2x 1 1 − cos2xdx 
2
2
 1  1 − 1 1  cos4x dx  1 
4
2
4
 1 x − 1 sin4x  C
4
8
2
1
1 1 − cos 2 2xdx
4
1 − 1 cos4x dx
2
2
 cos 2 2x sin 4 2xdx  
1 1  cos4x 1 1 − cos4x 2 dx
4
2
 1 1  cos4x1 − 2 cos4x  cos 2 4xdx
8
 1 1  cos4x − 2 cos4x − 2 cos 2 4x  cos 2 4x − cos 3 4xdx
8
 1 1 − cos4x − cos 2 4x − cos 3 4xdx
8
 1  1 − cos4x − 1 1  cos8x dx −  1 1 − sin 2 4xd 1 sin4x
2
4
2
8
 1 1 x − 1 sin4x − 1 sin8x − 1 sin4x  1 1 sin 3 4x
C
8 2
4
16
8
8 3
2. Consider the integrals of the form:
 tan m x sec n xdx
a. n is an even positive numbers.
 tan m x sec n xdx   tan m x sec n−2 x sec 2 xdx   tan m x1  tan 2  n−2/2 dtan x
b. m is an odd positive numbers.
 tan m x sec n xdx   tan m−1 x sec n−1 xtan x sec xdx  sec 2 − 1 m−1/2 sec n−1 xdsec x
Example Evaluate
1
 tan 4 x sec 4 xdx
2
 tan 5 x sec 2 xdx
(1) u  tan x, du  sec 2 xdx,
 tan 4 x sec 4 xdx

 tan 4 x1  tan 2 xdtan x   u 4 1  u 2 du
 1 u 5  1 u 7  C  1 tan 5 x  1 tan 7 x  C
7
5
7
5
(2) u  sec x, du  sec x tan xdx
 tan 5 x sec 2 xdx  sec 2 − 1 2 sec 2 xdsec x  u 4 − 2u 2  1u 2 du
 u 6 − 2u 4  u 2 du  1 u 7 − 2 u 5  1 u 3  C
7
5
3
2
1
1
7
5
3
sec x − sec x  sec x  C

5
7
3
Example Evaluate  sec xdx
x  tan x dx  
 sec xdx   sec x sec
sec x  tan x

2

u  sec x  tan x,
sec 2 x  tan x sec x dx,
sec x  tan x
du  sec x tan x  sec 2 xdx
1 du  ln|u|  C  ln|sec x  tan x|  C
u