Answer
$\frac{cos^{3}x}{3}-cos x+C$
Work Step by Step
$\int sin^{3}xdx= \int(1-cos^{2}x)sin x dx$
Plug in: t= cos x, so that $dx= -\frac{1}{sinx}dt$
Therefore, $\int sin^{3}xdx= \int(1-t^{2})sinx\times\frac{-1}{sinx}dt$
$=\int (t^{2}-1)dt= \int t^{2}dt-\int 1dt$
$=\frac{t^{3}}{3}-t+C$
Undoing substitution, we obtain
$\int sin^{3}xdx=\frac{cos^{3}x}{3}-cos x+C$
