#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 16 Maths Textbook Solution.

$\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+C$

Given:

$\int\left(\frac{1}{7+5 \cos x}\right) d x$

Hint:

You must know about the derivation of tan x and using $\int \frac{1}{1+x^{2}} d x$

Explanation:

Let $\mathrm{I}=\int \frac{1}{7+5 \cos x} d x$

$\text { Put } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

\begin{aligned} \therefore I &=\int \frac{1}{7+5\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{7\left(1+\tan ^{2} \frac{x}{2}\right)+5-5 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{12+2 \tan ^{2} \frac{x}{2}} \\ &=\frac{1}{2} \int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}+6} d x \\ &=\frac{1}{2} \int \frac{2 d t}{t^{2}+(\sqrt{6})^{2}} \end{aligned}    $\text { [ Put } \left.\tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t\right]$

\begin{aligned} &=\int \frac{d t}{t^{2}+(\sqrt{6})^{2}} \\ &=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{t}{\sqrt{6}}\right)+C \ldots \ldots \ldots \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \\ &=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{6}}\right)+C \end{aligned}