Evaluate the following integrals:
Question:

Evaluate the following integrals:

$\int \frac{1}{2+\sin x+\cos x} d x$

Solution:

Given I $=\int \frac{1}{2+\sin x+\cos x} d x$

We know that $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$ and $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

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

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

Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$ and putting $\tan x / 2=t$ and $\sec ^{2} x / 2 d x=2 d t$,

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{2+2 \tan ^{2} \frac{x}{2}-2 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x=\int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}-2 \tan \frac{x}{2}+3} d x$

$=\int \frac{2 d t}{t^{2}-2 t+3}$

$=2 \int \frac{1}{(t+1)^{2}+(\sqrt{2})^{2}} d t$

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

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

$=\sqrt{2} \tan ^{-1}\left(\frac{\tan \frac{x}{2}+1}{\sqrt{2}}\right)$

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

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