Question

Find
$\left(i\right)\int {\cos }^{2}xdx$
$\left(ii\right)\int \sin 2x\cos 3xdx$
$\left(iii\right)\int {\sin }^{3}xdx$

Answer 1

$\left(i\right)\int {\cos }^{2}xdx$
As we know
$2{\cos }^{2}x=\cos 2x+1$
$=\int \left(\frac{\cos 2x+1}{2}\right)dx$
$=\frac{1}{2}\int (\cos 2x+1)dx$
$=\frac{1}{2}[\int \cos 2xdx+\int dx]$
$=\frac{1}{2}[\frac{\sin 2x}{2}+x]+C$
$=\frac{\sin 2x}{4}+\frac{x}{2}+C$
Where $C$ is constant of integration.

$\left(ii\right)\int \sin 2x\cos 3xdx$
Multiply and divide by $2$
$=\frac{1}{2}\int 2\sin 2x\cos 3xdx$
$=\frac{1}{2}\int \left[\sin \right(2x+3x)+\sin (2x-3x\left)\right]dx$
$[\because 2\sin A\cos B=\sin (A+B)+\sin (A-B\left)\right]$
$=\frac{1}{2}\int \left[\sin \right(5x)+\sin (-x\left)\right]dx$
$=\frac{1}{2}\int [\sin 5x-\sin x]dx$
$=\frac{1}{2}\int \sin 5xdx-\frac{1}{2}\int \sin xdx$
​​​​​​​$=\frac{1}{2}[-\frac{1}{5}\cos 5x-(-\cos x\left)\right]+C$
​​​​​​​$=-\frac{1}{10}\cos 5x+\frac{1}{2}\cos x+C$

Where $C$ is constant of integration.

$\left(iii\right)\int {\sin }^{3}xdx$
Using $\sin 3x=3\sin x-4{\sin }^{3}x$
$=\int \frac{3\sin x-\sin 3x}{4}dx$
$=\frac{1}{4}\int (3\sin x-\sin 3x)dx$
$=\frac{1}{4}\left[3\right(-\cos x)-(-\frac{1}{3}\cos 3x)]+C$
$=\frac{1}{4}[-3\cos x+\frac{1}{3}\cos 3x]+C$
$=-\frac{3}{4}\cos x+\frac{1}{12}\cos 3x+C$

​​​​​​​Where $C$ is constant of integration.