## University Calculus: Early Transcendentals (3rd Edition)

$-\frac{1}{4}cos^{4}x +C$
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 $\int cos^{3}xsinx dx=-\frac{1}{4}cos^{4}x +C$