#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 30 Maths Textbook Solution.

$x-\tan x+C$

Given:

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

Hint:

Using $\int \tan ^{2} x d x$

Explanation:

Let $\mathrm{I}=\int \frac{\cos 2 x-1}{\cos 2 x+1} d x$

$=\int \frac{\cos ^{2} x-\sin ^{2} x-1}{\cos ^{2} x-\sin ^{2} x+1} d x$                                                        $\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right]$

\begin{aligned} &=\int \frac{\left(\cos ^{2} x-1\right)-\sin ^{2} x}{\cos ^{2} x+\left(1-\sin ^{2} x\right)} d x \\ &=\int \frac{-\sin ^{2} x-\sin ^{2} x}{\cos ^{2} x+\cos ^{2} x} d x \end{aligned}                                                      $\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow 1-\sin ^{2} x=\cos ^{2} x \end{array}\right]$

\begin{aligned} &=\int \frac{-2 \sin ^{2} x}{2 \cos ^{2} x} d x \\ &=-\int \tan ^{2} x d x \end{aligned}                                                                              $\left[\because \tan x=\frac{\sin x}{\cos x}\right]$

\begin{aligned} &=-\int\left(\sec ^{2} x-1\right) d x \\ &=\int\left(1-\sec ^{2} x\right) d x \\ &=\int 1 d x-\int \sec ^{2} x d x \\ &=x-\tan x+C \end{aligned}

Hence,$\int \frac{\cos 2 x-1}{\cos 2 x+1} d x=x-\tan x+C$