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

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

Answers (1)

Answer:

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}

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