# What is the integral of int sin^2 (x).cos^2 (x) dx ?

Jun 19, 2016

$\int {\sin}^{2} x {\cos}^{2} x \mathrm{dx} = \frac{x}{8} - \frac{\sin 4 x}{32} + c$

#### Explanation:

As $\sin 2 x = 2 \sin x \cos x$

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

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

= $\frac{1}{4} \int \frac{1 - \cos 4 x}{2} \mathrm{dx}$

= $\frac{x}{8} - \frac{1}{8} \int \cos 4 x \mathrm{dx}$

= $\frac{x}{8} - \frac{1}{8} \times \frac{\sin 4 x}{4} + c$

= $\frac{x}{8} - \frac{\sin 4 x}{32} + c$