∫x(sinx+cosx)² dx
=∫x(1+sin2x) dx
=∫x dx+∫xsin2x dx
=1/2*x²-1/2*∫x d(cos2x)
=1/2*x²-1/2*xcos2x+1/2*∫cos2x dx
=1/2*x²-1/2*xcos2x+1/4*∫cos2x d(2x)
=1/2*x²-1/2*xcos2x+1/4*sin2x
然后将积分上下限带进去计算即可
y=sin^4x+cos^4x
y'=4sin³x*(sinx)'+4cos³x*(cosx)'
=4sin³x*cosx-4cos³x*sinx
=4sinxcosx(sin²x-cos²x)
=2sin2x*(-cos2x)
=-2sin2x*cos2x
=-cos4x就最后一步y =-cos4x 应该是y =-sin4x 吧，我懂了，谢额，是的，汗一个！！！光犯一些低级错误……