# Evaluate intcos^5x dx using u-substitution or tabular integration?

Apr 30, 2018

$\sin x - \frac{2}{3} {\sin}^{3} x + \frac{1}{5} {\sin}^{5} x + C$

#### Explanation:

We will use u-substitution and a trigonometric identity.

Recall that ${\cos}^{2} x + {\sin}^{2} x = 1$. Rearranging, it follows that ${\cos}^{2} x = 1 - {\sin}^{2} x$. Note that we can use this fact to change the integral as follows:

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

Make a u-substitution by letting $u = \sin x$. Then $\mathrm{du} = \cos x \mathrm{dx}$. Our integral becomes:

$\int {\left(1 - {u}^{2}\right)}^{2} \mathrm{du} = \int \left(1 - 2 {u}^{2} + {u}^{4}\right) \mathrm{du}$
$= u - \frac{2}{3} {u}^{3} + \frac{1}{5} {u}^{5} + C$
$= \sin x - \frac{2}{3} {\sin}^{3} x + \frac{1}{5} {\sin}^{5} x + C$