# Integral of cos^5x*sin³xdx ?

May 5, 2018

$- \frac{1}{6} {\cos}^{6} x + \frac{1}{8} {\cos}^{8} x + C$

#### Explanation:

We wish to solve the following: $\int \left({\cos}^{5} x {\sin}^{3} x\right) \mathrm{dx}$. Since $\cos x$ is the derivative of $\sin x$, this looks ripe for a u-substitution. However, we'd like there to be only one instance of $\mathrm{du}$, so we need to figure out a way to transform the integral so that we only have a single power of $\sin$ or $\cos$.

Luckily, this is easy. Recall the identity ${\sin}^{2} x + {\cos}^{2} x = 1$. From this, it follows that ${\sin}^{2} x = 1 - {\cos}^{2} x$. Using this fact, we can make the following changes:

$\int \left({\cos}^{5} x {\sin}^{3} x\right) \mathrm{dx}$
$= \int \left({\cos}^{5} x {\sin}^{2} x \sin x\right) \mathrm{dx}$
$= \int \left(\left({\cos}^{5} x\right) \left(1 - {\cos}^{2} x\right) \left(\sin x\right)\right) \mathrm{dx}$

Now, let $u = \cos x$. Then $\mathrm{du} = - \sin x$. This gives

$- \int \left(\left({u}^{5}\right) \left(1 - {u}^{2}\right)\right) \mathrm{du}$
$= - \int \left({u}^{5} - {u}^{7}\right) \mathrm{du}$
$= - \left(\frac{1}{6} {u}^{6} - \frac{1}{8} {u}^{8}\right) + C$
$= - \frac{1}{6} {\cos}^{6} x + \frac{1}{8} {\cos}^{8} x + C$