请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
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![请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?](/uploads/image/z/12890014-70-4.jpg?t=%E8%AF%B7%E9%97%AE%E2%88%ABsin%28u%2F2%29%2Asin%28u%2F2%29du%E6%88%96%E2%88%ABsin%5E2+%28u%2F2%29du%E6%80%8E%E4%B9%88%E8%A7%A3%E5%95%8A%3F)
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
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
∫sin^2 (u/2)du=∫(1-cosu)/2 du=u/2-(sinu)/2