请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
问题描述：
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
答
cosu=1-2[sin(u/2)]^2
所以[sin(u/2)]^2=(1-cosu)/2
所以∫sin^2 (u/2)du
=∫(1-cosu)/2du
=∫1/2du-∫(cosu)/2du
=u/2-(1/2)*sinu+C