#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 2 Maths Textbook Solution.

$\frac{1}{2} \log \left(\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right)+C$

Given:

$\int \frac{1}{\cos x+\sqrt{3} \sin x} d x$

Hint:

You must know about the $\int \cos e c x d x$

Explanation:

Let $I=\int \frac{1}{2\left(\sin x \cdot \frac{\sqrt{8}}{2}+\cos x \cdot \frac{1}{2}\right)} d x$

$=\frac{1}{2} \int \frac{d x}{\sin x \cos _{6}^{\pi}+\cos x \sin \frac{\pi}{6}}$                                $\left[\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} ; \sin \frac{\pi}{6}=\frac{1}{2}\right]$

$=\frac{1}{2} \int \frac{d x}{\sin \left(x+\frac{\pi}{6}\right)}$                                                    $[\sin (A+B)=\sin A \cos B+\cos A \sin B]$

$=\frac{1}{2} \int \operatorname{cosec}\left(x+\frac{\pi}{6}\right) d x$                                                   $\left[\because \sin x=\frac{1}{\cos \theta c x}\right]$

$=\frac{-1}{2} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{6}\right)+\cot \left(x+\frac{\pi}{6}\right)\right)+C$        $\left[\int \operatorname{cosec}(x) d x=-\log (\operatorname{cosec} x+\cot x)+c\right]$

Now,

\begin{aligned} \cos e c x+\cot x &=\frac{1}{\sin x}+\frac{\cos x}{\sin x} \\ &=\frac{1+\cos x}{\sin x} \\ &=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos _{2}^{x}} \end{aligned}                                                     $\left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta ; \sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]$

$=\cot \frac{x}{2}$

$\therefore \text { From } E q \cdot(i)$

\begin{aligned} &I=\frac{-1}{2} \log \left(\cot \left(\frac{x+\frac{\pi}{6}}{2}\right)\right)+C \\ &=\frac{1}{2} \log \left[\cot \left(\frac{x}{2}+\frac{\pi}{12}\right)\right]^{-1}+C \\ &=\frac{1}{2} \log \left(\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right)+C \end{aligned}