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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 xdcos 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 x1 − sin 2 x m−1/2 dsin x b. Both n and m are positive even numbers: Use the identities: sin 2 x 1 1 − cos2x, and cos 2 x 1 1 cos2x 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 x1 − cos 2 xdcos 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 x1 − sin x dsin 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 2xdx 1 cos 2 x sin 2 xdx 1 1 cos2x 1 1 − cos2xdx 2 2 1 1 − 1 1 cos4x dx 1 4 2 4 1 x − 1 sin4x C 4 8 2 1 1 1 − cos 2 2xdx 4 1 − 1 cos4x dx 2 2 cos 2 2x sin 4 2xdx 1 1 cos4x 1 1 − cos4x 2 dx 4 2 1 1 cos4x1 − 2 cos4x cos 2 4xdx 8 1 1 cos4x − 2 cos4x − 2 cos 2 4x cos 2 4x − cos 3 4xdx 8 1 1 − cos4x − cos 2 4x − cos 3 4xdx 8 1 1 − cos4x − 1 1 cos8x dx − 1 1 − sin 2 4xd 1 sin4x 2 4 2 8 1 1 x − 1 sin4x − 1 sin8x − 1 sin4x 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 x1 tan 2 n−2/2 dtan x b. m is an odd positive numbers. tan m x sec n xdx tan m−1 x sec n−1 xtan x sec xdx sec 2 − 1 m−1/2 sec n−1 xdsec 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 x1 tan 2 xdtan 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 xdsec x u 4 − 2u 2 1u 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 xdx 1 du ln|u| C ln|sec x tan x| C u