# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{2}{2+\sin 2 x} d x$

Solution:

Given $I=\int \frac{2}{2+\sin 2 x} d x$

We know that $\sin 2 x=2 \sin x \cos x$

$\Rightarrow \int \frac{2}{2+\sin 2 x} d x=\int \frac{2}{2+2 \sin x \cos x} d x$

$=\int \frac{1}{1+\sin x \cos x} d x$

Dividing the numerator and denominator by $\cos ^{2} x$,

$\Rightarrow \int \frac{1}{1+\sin x \cos x} d x=\int \frac{\sec ^{2} x}{\sec ^{2} x+\tan x} d x$

Replacing $\sec ^{2} x$ in denominator by $1+\tan ^{2} x$,

$\Rightarrow \int \frac{\sec ^{2} x}{\sec ^{2} x+\tan x} d x=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+\tan x} d x$

Putting $\tan x=t$ so that $\sec ^{2} x d x=d t$,

$\Rightarrow \int \frac{\sec ^{2} x}{\tan ^{2} x+\tan x+1} d x=\int \frac{d t}{t^{2}+t+1}$

$=\int \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$

We know that $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

$\Rightarrow \int \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{\mathrm{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\mathrm{c}$

$=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{t}+1}{\sqrt{3}}\right)+\mathrm{c}$

$\therefore \mathrm{I}=\int \frac{2}{2+\sin 2 \mathrm{x}} \mathrm{dx}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{t}+1}{\sqrt{3}}\right)+\mathrm{c}$