Answer
$\frac{1}{3}$$sin^{3}$x-$\frac{1}{5}$$sin^{5}$x+C
Work Step by Step
$\int$$sin^{2}$x$cos^{3}$xdx
=$\int$$sin^{2}$x$cos^{2}$xcosxdx, split the odd power
=$\int$$sin^{2}$x(1-$sin^{2}$x)cosxdx, make $cos^{2}$=(1-$sin^{2}$x)
Now use u-sub:
u=sinx; du=cosxdx
=$\int$$u^{2}$(1-$u^{2}$)du
=$\int$($u^{2}$-$u^{4}$)du
=$\frac{1}{3}$$u^{3}$x-$\frac{1}{5}$$u^{5}$x+C
=$\frac{1}{3}$$sin^{3}$x-$\frac{1}{5}$$sin^{5}$x+C
