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#### Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 3 maths textbook solution.

Answer : $\frac{1}{\sqrt{3}} \log \left|\frac{\tan \frac{x}{2}-2-\sqrt{3}}{\tan \frac{x}{2}-2+\sqrt{3}}\right|+c$

Hint : To solve this equation we have to convert sin into tan form

Given : $\int \frac{1}{1-2 \sin x} d x$

Solution : $I=\int \frac{1}{1-2 \sin x} d x$

Using the formula : $\left[\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]$

\begin{aligned} &I=\int \frac{1}{1-\frac{4 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-4 \tan \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-4+1} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-(\sqrt{3})^{2}} d x \end{aligned}

\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-(\sqrt{3})^{2}} d x \\ &{\left[\begin{array}{l} t=\tan \frac{x}{2} \\ d t=\sec ^{2} \frac{x}{2} \times \frac{1}{2} d x \\ 2 d t=\left(\sec ^{2} \frac{x}{2}\right) d x \end{array}\right]} \end{aligned}

\begin{aligned} &I=\int \frac{2 d t}{(t-2)^{2}-(\sqrt{3})^{2}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right] \\ &I=\frac{2}{2 \sqrt{3}} \log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|+c \end{aligned}

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