# How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dx ?

Sep 21, 2014

The answer is $- \frac{{\cos}^{3} x}{3} + \frac{{\cos}^{5} x}{5} + C$.

The trick with sinusoidal powers is to use identities so that you can have $\sin x$ or $\cos x$ with a power of 1 and use substitution.

In this case, it is easier to get $\sin x$ to a power of 1 using ${\sin}^{2} x = 1 - {\cos}^{2} x$.

$\int {\sin}^{3} x \cdot {\cos}^{2} x \mathrm{dx}$
$= \int \sin x \left(1 - {\cos}^{2} x\right) {\cos}^{2} x \mathrm{dx}$
$= \int \sin x \left({\cos}^{2} x - {\cos}^{4} x\right) \mathrm{dx}$
$= \int \sin x {\cos}^{2} x \mathrm{dx} - \int \sin x {\cos}^{4} x \mathrm{dx}$

Now it is a matter of using substitution:

$u = \cos x$
$\mathrm{du} = - \sin x \mathrm{dx}$

$\int \sin x {\cos}^{2} x \mathrm{dx} - \int \sin x {\cos}^{4} x \mathrm{dx}$
$= \int - {u}^{2} \mathrm{du} + \int {u}^{4} \mathrm{du}$
$= - \frac{{u}^{3}}{3} + \frac{{u}^{5}}{5} + C$
$= - \frac{{\cos}^{3} x}{3} + \frac{{\cos}^{5} x}{5} + C$