Question

请问这一题为什么两种解法的答案不同呢？

第二个是没有用二倍的性质来做

Answer 1

如图所示

Answer 2

∫x.sin2x dx

=-(1/2)∫x.dcos2x

=-(1/2)xcos2x +(1/2)∫cos2x dx

=-(1/2)xcos2x +(1/4)sin2x +C

∫(0->π) x√[(cosx)^2-(cosx)^4] dx

=∫(0->π) x√(sinx)^2.(cosx)^2] dx

=∫(0->π/2) x.(sinx)(cosx) dx -∫(π/2->π) x.(sinx)(cosx) dx

=(1/2)∫(0->π/2) xsin2x dx -(1/2)∫(π/2->π) x.sin2x dx

=(1/2)[-(1/2)xcos2x +(1/4)sin2x]|(0->π/2) -(1/2)[-(1/2)xcos2x +(1/4)sin2x]|(π/2->π)

=(1/2)(π/4 ) +(1/2)(π/4)

=π/4

Answer 3

<img>b90e7bec54e736d1997204d495504fc2d462698d</img>

追答