Question

这个积分怎么求，谢谢。

Answer 1

0<x<π则sinx>0
所以原式=∫(0,π)x√[sin²x(1-sin²x)dx
=∫(0,π)x√(sin²xcos²x)dx
=∫(0,π)xsinx|cosx|dx
=∫(0,π/2)xsinxcosxdx+∫(π/2,π)xsinx（-cosx）dx
=∫(0,π/2)x*1/2*sin2xdx-∫(π/2,π)x*1/2*sin2xdx
=1/4*∫(0,π/2)xsin2xd(2x)-1/4*∫(π/2,π)xsin2xd(2x)
=-1/4*∫(0,π/2)xdcos2xd(2x)+1/4*∫(π/2,π)xdcos2x
=-1/4*∫(0,π/2)xdcos2xd(2x)+1/4*∫(π/2,π)xdcos2x
接下去分别算
∫(0,π/2)xdcos2x
=xcos2x(0,π/2)-∫(0,π/2)cos2xdx
=xcos2x(0,π/2)-1/2*sin2x(0,π/2)
=-π/2
同理∫(π/2,π)xdcos2x=3π/2
所以原式=π/8+3π/8=π/2