∫[(sin^2)x]dx求不定积分
问题描述：
∫[(sin^2)x]dx求不定积分
答
∫[(sin^2)x]dx
=1/2*∫(1-cos(2x))dx
=1/2*(x-1/2*sin(2x))+C
=x/2-sin(2x)/4+C