# What is the net area between f(x) = cos^2xsin^2x  and the x-axis over x in [0, 3pi ]?

Mar 12, 2017

The answer is $= \frac{3}{8} \pi = 1.178 {u}^{2}$

#### Explanation:

We start, by simplifying

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

and

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

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

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

Therefore,

$f \left(x\right) = {\cos}^{2} x {\sin}^{2} x$

$= \frac{1}{4} {\sin}^{2} 2 x$

$= \frac{1}{4} \cdot \frac{1}{2} \cdot \left(1 - \cos 4 x\right)$

$= \frac{1}{8} \left(1 - \cos 4 x\right)$

The area is

$= \frac{1}{8} {\int}_{0}^{3 \pi} \left(1 - \cos 4 x\right) \mathrm{dx}$

$= \frac{1}{8} {\left[x - \frac{1}{4} \sin 4 x\right]}_{0}^{3 \pi}$

$= \frac{1}{8} \left(3 \pi - \frac{1}{4} \sin 12 \pi - 0 - 0\right)$

$= \frac{3}{8} \pi = 1.178$