How do you use Integration by Substitution to find intcos^3(x)*sin(x)dx?

Sep 8, 2014

By substitution,
$\int {\cos}^{3} x \sin x \mathrm{dx} = - \frac{1}{4} {\cos}^{4} x + C$

Let $u = \cos x$.
By taking the derivative,
$\frac{\mathrm{du}}{\mathrm{dx}} = - \sin x$
By taking the reciprocal,
$\frac{\mathrm{dx}}{\mathrm{du}} = \frac{1}{- \sin x}$
By multiplying by $\mathrm{du}$,
$\mathrm{dx} = \frac{\mathrm{du}}{- \sin x}$

By rewriting the integral in terms of $u$,
intcos^3xsinx dx =int u^3 sinx cdot {du}/{-sinx}
by cancelling out $\sin x$'s,
$= - \int {u}^{3} \mathrm{du}$
by Power Rule,
$= - {u}^{4} / 4 + C$
by putting $u=cosx$ back in,
$= - \frac{1}{4} {\cos}^{4} x + C$