# What is the integral of int sin^3xcos^2x dx from 0 to pi/2?

May 6, 2018

$\frac{2}{15}$

#### Explanation:

Rewrite the integrand as ${\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} x {\cos}^{2} x \sin x \mathrm{dx}$

Recalling the identity ${\sin}^{2} x = 1 - {\cos}^{2} x ,$ we get

${\int}_{0}^{\frac{\pi}{2}} \left(1 - {\cos}^{2} x\right) {\cos}^{2} x \sin x \mathrm{dx}$

We can now solve this with a simple substitution.

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

Calculate the new bounds:

Upper: $u = \cos \left(\frac{\pi}{2}\right) = 0$
Lower: $u = \cos 0 = 1$

$- {\int}_{1}^{0} {u}^{2} \left(1 - {u}^{2}\right) \mathrm{du} = - {\int}_{1}^{0} \left({u}^{2} - {u}^{4}\right) \mathrm{du}$

$= - \left(\frac{1}{3} {u}^{3} - \frac{1}{5} {u}^{5}\right) {|}_{1}^{0}$
$- \left(- \frac{1}{3} + \frac{1}{5}\right) = \frac{2}{15}$