# Question 642d1

Jan 17, 2018

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

#### Explanation:

Using the trigonometric identity:

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

we have that:

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

so:

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

Use now:

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

to have:

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

$\int {\cos}^{2} x {\sin}^{2} x \mathrm{dx} = \frac{1}{8} \int \mathrm{dx} - \frac{1}{32} \int \cos 4 x d \left(4 x\right)$

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

Jan 17, 2018

$\int {\cos}^{2} x {\sin}^{2} x \mathrm{dx} = \frac{1}{8} x - \frac{1}{32} \sin \left(4 x\right) + \text{C}$

#### Explanation:

Given: $\int {\cos}^{2} x \sin {x}^{2} x \mathrm{dx}$

The trick is to use a number of identities:

Use the half angle identities to rewrite the integral:

color(blue)(sin^2x=1/2(1-cos(2x))

color(blue)(cos^2x=1/2(1+cos(2x))

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

Take out the constants: $\left(\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\right)$

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

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

Use the identity color(blue)(sin^2x+cos^2x=1=>1-cos^2x=sin^2x

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

Use the identity: color(blue)(sin^2x=(1-cos(2x))/2#

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

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

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

Integrate each term:

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

$= \frac{1}{8} \left[x - \frac{1}{4} \sin \left(4 x\right)\right] + \text{C}$

$= \frac{1}{8} x - \frac{1}{32} \sin \left(4 x\right) + \text{C}$