∫sin²xcos³x dx

∫sin²xcos³x dx
∫sin²xcos³x dx

∫sin²xcos³x dx
∫sin^2xcos^3xdx = ∫sin^2x (1-sin^2x)dsinx= ∫sin^2x-sin^4x dx = (1/3) sin^3x-(1/5)sin^5x+C
不是让你求助我吗.

∫sin²xcos³xdx
=∫sin²xcos²xdsinx
=∫sin²x(1-sin²x)dsinx
=∫sin²xdsinx-∫sin^4xdsinx
=1/3sin³x-1/5sin^5x+C