题目

求定积分 上限π 下限0 [(xsin x)^2] dx
求定积分 上限π 下限0 [(xsin x)^2] dx

提问时间：2020-08-07

答案

I= ∫(0->π) (xsinx)^2 dx
= (1/2)∫(0->π) x^2(1-cos2x) dx
= (1/2)[x^3/3](0->π) - (1/4)∫(0->π) x^2 .dsin2x
= π^3/6 - (1/4)[x^2sin2x](0->π) + (1/2)∫(0->π) xsin2x dx
= π^3/6 - (1/4)∫(0->π) xdcos2x
= π^3/6 - (1/4)[xcos2x](0->π) + (1/4)∫(0->π) cos2x dx
= π^3/6 - π/4 + [sin2x/2](0->π)
=π^3/6 - π/4