Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 62  Maths Textbook Solution.

$\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}$

Hint:

To solve the given statement we will take $\cos 2 x \operatorname{as}\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)$

Given:

$\int \frac{1}{2-3 \cos 2 x} d x$

Solution:

$I=\int \frac{1}{2-3\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)} d x$

$I=\int \frac{1+\tan ^{2} x}{2\left(1+\tan ^{2} x\right)-3\left(1-\tan ^{2} x\right)} d x$

$I=\int \frac{\sec ^{2} x}{-1+\operatorname{stan}^{2} x} d x$

$I=\int \frac{\sec ^{2} x}{5\left(\tan ^{2} x-\frac{1}{5}\right)} d x$

$I=\int \frac{1}{5} \frac{d t}{t^{2}-\frac{1}{5}}$                                        $\left[\because \tan x=t, d t=\sec ^{2} x d x\right]$

$I=\frac{1}{5} \int \frac{d t}{t^{2}-\left(\frac{1}{\sqrt{5}}\right)^{2}}$

$I=\frac{1}{5} \frac{1}{2\left(\frac{1}{\sqrt{5}}\right)} \ln \left|\frac{t-\left(\frac{1}{\sqrt{5}}\right)}{t+\left(\frac{1}{\sqrt{5}}\right)}\right|+c$

$I=\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}$