# How to solve ∫ (sin^4x)(cos^2x) dx ?

Sep 30, 2015

$\int {\sin}^{4} x \cdot {\cos}^{2} x \mathrm{dx} = \frac{1}{16} \cdot \left(x - \sin \frac{4 x}{4} - \frac{{\sin}^{3} \left(2 x\right)}{3}\right) + c$

#### Explanation:

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

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

We have

$\cos 2 x = 1 - 2 {\sin}^{2} x$

${\sin}^{2} x = \frac{1 - \cos 2 x}{2}$

and

$\sin 2 x = 2 \sin x \cdot \cos x$

The integral becomes

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

$= \frac{1}{8} \left(\int \left(1 - \cos 2 x\right) \cdot {\sin}^{2} \left(2 x\right)\right) \mathrm{dx}$

=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx

$= \frac{1}{8} \left(\int \left(\frac{1}{2}\right) \cdot 2 {\sin}^{2} \left(2 x\right) \mathrm{dx} - \int \frac{1}{2} \cdot 2 \cos 2 x \cdot {\sin}^{2} \left(2 x\right) \mathrm{dx}\right)$

$= \frac{1}{8} \left(\int \left(\frac{1}{2}\right) \cdot \left(1 - \cos \left(4 x\right)\right) \mathrm{dx} - \int \frac{1}{2} \cdot {\sin}^{2} \left(2 x\right) d \left(\sin 2 x\right)\right)$

$= \frac{1}{16} \left(\int \left(1 - \cos \left(4 x\right)\right) \mathrm{dx} - \int {\sin}^{2} \left(2 x\right) d \left(\sin 2 x\right)\right)$

$d \frac{\sin 2 x}{\mathrm{dx}} = 2 \cos 2 x$

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

$= \frac{1}{16} \left(x - \sin \frac{4 x}{4} - {\sin}^{3} \frac{2 x}{3}\right) + c$