Since sin²x=1-cos²x, the given function can be rewritten as follows:
∫sin²xcos³xdx=∫(1-cos²x)cos³xdx=∫cos³xdx - ∫cos^5(x)dx
From trig identies such as, 4cos³x=cos3x-3cosx, and 16cos^5(x)=cos5x+5cos3x+10cosx, the function above can be rewritten as follows:
∫sin²xcos³x=(1/4)∫(cos3x+3cosx)dx - (1/16)∫(cos5x+5cos3x+10cosx)dx
From anti-derivative rules such as, ∫cos(ax+b)dx=(1/a)sin(ax+b)+C, and ∫cosxdx=sinx+C, the function above can be rewritten as follows:
∫sin²xcos³xdx=(1/4)((1/3)sin3x+3sinx+C) - (1/16)((1/5)sin5x+(5/3)sin3x+10sinx+C)
=(1/16)(-(1/5)sin5x-(1/3)sin3x+2sinx)+C=(-1/16)((1/5)sin5x+(1/3)sin3x-2sinx)+C
Therefore, the answer is: ∫sin²xcos³xdx=(-1/16)((1/5)sin5x+(1/3)sin3x-2sinx) + C