Answer
$-\frac{1}{4}cos^{4}x +C$
Work Step by Step
Plug in: t= cos x so that $dx= \frac{dt}{-sin x}$ in the given integral.
Then we have
$\int cos^{3}xsinx dx= \int t^{3}sinx\frac{dt}{-sinx}$
$= -\int t^{3}dt= -\frac{t^{4}}{4}+C$
Undoing substitution, we get
$\int cos^{3}xsinx dx=-\frac{1}{4}cos^{4}x +C$
