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

Sep 19, 2014

$= - \frac{{\cos}^{3} x}{3} + \frac{{\cos}^{5} x}{5} + c$, where $c$ is a constant

Explanation :

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

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

from trigonometric identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, we get ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$

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

let's $\cos \left(x\right) = t$, $\implies - \sin \left(x\right) \mathrm{dx} = \mathrm{dt}$

$= \int \left(1 - {t}^{2}\right) \cdot {t}^{2} \left(- \mathrm{dt}\right)$

$= - \int \left(1 - {t}^{2}\right) \cdot {t}^{2} \mathrm{dt}$

$= - \int \left({t}^{2} - {t}^{4}\right) \mathrm{dt}$

$= - \int {t}^{2} \mathrm{dt} + \int {t}^{4} \mathrm{dt}$

$= - {t}^{3} / 3 + {t}^{5} / 5 + c$, where $c$ is a constant

substituting $t$ back, yields

$= - \frac{{\cos}^{3} x}{3} + \frac{{\cos}^{5} x}{5} + c$, where $c$ is a constant