解法一：第一换元法（凑微分法）
∫sin³xcos³xdx
=∫sin²xcos³xsinxdx
=-∫(1-cos²x)cos³xdcosx
=∫cos^(5)xdcosx-∫cos³xdcosx
=cos^(6)x/6-cos^(4)x/4+C
解法二：三倍角公式sin3x=3sinx-4sin³x
∫sin³xcos³xdx
=∫(sinxcosx)³dx
=∫(sin2x/2)³dx
=∫sin³2xdx/8
=∫(3sin2x/4-sin6x/4)dx/8
=3∫sin2xdx/32-∫sin6xdx/32
=3∫sin2xd2x/64-∫sin6xd6x/192
=-3cos2x/64+cos6x/192+C