不定积分题目

∫f'(cosx) dx= cosx +C
f'(cosx) = -sinx
∫f'(cosx) dcosx = ∫ -sinx dcosx
f(cosx)
=∫ (sinx)^2 dx
=(1/2) ∫ (1-cos2x) dx
=(1/2)[ x -(1/2)sin2x] + C