Section 6.2--Trigonometric Integrals and Substitutions

(cos2 x)2dx = Z 1+cos2x 2 2 dx = 1 4 Z (1+cos2x)2dx = 1 4 Z (1+2cos2x +cos2 2x)dx = 1 4 Z 1+2cos2x+ 1 2 (1+cos4x) dx = 1 4 Z 3 2 +2cos2x+ 1 2 cos4x dx = 1 4 · 3 2 x+ 1 4 ·2· 1 2 sin2x + 1 4 · 1 2 · 1 4 sin4x+C = 3 8 x+ 1 4 sin2x+ 1 32 sin4x+C Case B: Integrals of type Z tanm xsecn xdx Midterms where m and n are nonnegative integers. METHOD ... ................
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