Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 24 maths

$I=\log \sqrt{3}$

Given:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin 2 x} d x$

Hint:

To solve this equation we convert sin into cosec
Explanation:

Let

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin 2 x} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \operatorname{cosec} 2 x \; d x \end{aligned}

\begin{aligned} &I=-\frac{1}{2}[\log |\operatorname{cosec} 2 x+\cot 2 x|]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=-\frac{1}{2}\left[\log \left(\operatorname{cosec} \frac{2 \pi}{3}+\cot \frac{2 \pi}{3}\right)\right]-\left[\log \left(\operatorname{cosec} \frac{\pi}{3}+\cot \frac{\pi}{3}\right)\right] \end{aligned}

\begin{aligned} &I=-\frac{1}{2}\left[\log \left|\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right|\right]-\left[\log \left|\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right|\right] \\\\ &I=-\frac{1}{2}\left[\log \frac{1}{\sqrt{3}}-\log \sqrt{3}\right] \end{aligned}

\begin{aligned} &I=-\frac{1}{2}[-\log \sqrt{3}-\log \sqrt{3}] \\\\ &I=-\frac{1}{2} \times-2 \log \sqrt{3} \\\\ &I=\log \sqrt{3} \end{aligned}