Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{1-2 \sin x} d x$

Solution:

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 $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c$

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

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

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

Leave a comment

None
Free Study Material