∫sin的二次方x乘以cos四次方的xdx怎么求,求过程,感谢. ∫sin^2xcos^4xdx=?
1个回答
∫sin²xcos^4xdx
=∫sin²xcos²xcos²xdx
=∫(1/4)sin²2xcos²xdx
=(1/4)∫(1/4)(1-cos4x)*(cos2x+1)dx
=(1/16)∫(cos2x+1-cos4xcos2x-cos4x)dx
=(1/16)[∫cos2xdx+∫dx-∫cos4xcos2xdx-∫cos4xdx]
=(1/16)[(1/2)sin2x+x-(1/2)∫cos4xd(sin2x)-(1/4)sin4x]
=(1/16)[(1/2)sin2x+x-(1/2)∫(1-2sin²2x)d(sin2x)-(1/4)sin4x]
=(1/16)[(1/2)sin2x+x-(1/2)sin2x-∫sin²2xd(sin2x)-(1/4)sin4x]
=(1/16)[(1/2)sin2x+x-(1/2)sin2x-(1/3)sin³3x-(1/4)sin4x]+C
=(1/32)[x-(1/3)sin³3x-(1/4)sin4x]+C
