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?

∫sin^2 (u/2)du=∫(1-cosu)/2 du=u/2-(sinu)/2

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