#### Need Solution for R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 19  Maths Textbook Solution.

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

Given:

$\int \frac{\sin x}{3+4 \cos ^{2} x} d x$

Hint:

You must know about the derivative of cos x and $\int \frac{1}{1+x^{2}} d x$

Explanation:

Let  $I=\int \frac{\sin x}{3+4 \cos ^{2} x} d x$

$=\int \frac{-d t}{3+4 t^{2}}$                                                                $[\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t]$

\begin{aligned} &=\frac{-1}{4} \int \frac{d t}{t^{2}+\frac{3}{4}} \\ =& \frac{-1}{4} \int \frac{d t}{(t)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ =& \frac{-1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{3}}{2}}\right)+c \ldots \ldots \int \frac{1}{x^{2}+a^{2}} d x=\tan ^{-1} \frac{x}{a}+c \\ =& \frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 t}{\sqrt{3}}\right)+C \\ =& \frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C \end{aligned}