Question

\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{1}{1 - 2 \sin x} dx \]

\[\text{ Putting sin x }= \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]

\[ \therefore I = \int\frac{1}{1 - 2 \left( \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)}dx\]

\[ = \int\frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{1 + \tan^2 \frac{x}{2} - 4 \tan \frac{x}{2}} dx\]

\[ = \int\frac{\sec^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} - 4 \tan \frac{x}{2}}dx\]

\[\text{ Putting   tan }\frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \text{ sec}^2 \left( \frac{x}{2} \right) dx = dt\]

\[ \Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right) \cdot dx = 2dt\]

\[ \therefore I = \int\frac{2}{t^2 - 4t + 1}dt\]

\[ = 2\int\frac{1}{t^2 - 4t + 4 - 4 + 1}dt\]

\[ = 2\int\frac{1}{\left( t - 2 \right)^2 - \left( \sqrt{3} \right)^2}dt\]

\[ = 2 \times \frac{1}{2\sqrt{3}} \text{ ln }\left| \frac{t - 2 - \sqrt{3}}{t - 2 + \sqrt{3}} \right| + C .......................\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]

\[ = \frac{1}{\sqrt{3}} \text{ ln } \left| \frac{\tan\frac{x}{2} - 2 - \sqrt{3}}{\tan\frac{x}{2} - 2 + \sqrt{3}} \right| + C .........................\left[ \because t = \tan \frac{x}{2} \right]\]