Evaluate the following integrals:
$\int \frac{1}{1-2 \sin x} d x$
Given $I=\int \frac{1}{1-2 \sin x} d x$
We know that $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan \frac{x}{2}}$
$\Rightarrow \int \frac{1}{1-2 \sin x} d x=\int \frac{1}{1-2\left(\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2}{2}}\right)} d x$
$=\int \frac{1+\tan ^{2} \frac{x}{2}}{1\left(1+\tan ^{2} \frac{x}{2}\right)-2\left(2 \tan \frac{x}{2}\right)} d x$
Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$,
$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{1\left(1+\tan ^{2} \frac{x}{2}\right)-2\left(2 \tan \frac{x}{2}\right)} d x=\int \frac{\sec ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-4 \tan \frac{x}{2}} d x$
Putting $\tan x / 2=t$ and $\sec ^{2}(x / 2) d x=2 d t$
$\Rightarrow \int \frac{\sec ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-4 \tan \frac{x}{2}} d x=\int \frac{2 d t}{1+t^{2}-4 t}$
$=2 \int \frac{1}{t^{2}-4 t+1} d t$
$=2 \int \frac{1}{(t-2)^{2}-(\sqrt{3})^{2}} d t$
We know that $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c$
$\Rightarrow 2 \int \frac{1}{(t-2)^{2}-(\sqrt{3})^{2}} d t=2\left(\frac{1}{2 \sqrt{3}}\right) \tan ^{-1}\left(\frac{t-2-\sqrt{3}}{t+2+\sqrt{3}}\right)+c$
$=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan x-(2+\sqrt{3})}{\tan x+(2+\sqrt{3})}\right)+c$
$\therefore \mathrm{I}=\int \frac{1}{1-2 \sin \mathrm{x}} \mathrm{dx}=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \mathrm{x}-(2+\sqrt{3})}{\tan \mathrm{x}+(2+\sqrt{3})}\right)+\mathrm{c}$
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