Question

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

Answer 1

\[\text{ Let I }= \int \frac{1}{1 - 2 \sin x}   \text{ dx }\]
\[\text{  Putting }\ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow I = \int\frac{1}{1 - 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \int \frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 1 + \tan^2 \frac{x}{2} \right) - 4 \tan \left( \frac{x}{2} \right)}dx\]
\[ = \int \frac{\text{ sec}^2 \left( \frac{x}{2} \right)}{\tan^2 \left( \frac{x}{2} \right) - 4 \tan \left( \frac{x}{2} \right) + 1} dx\]
\[\text{  Let tan} \left( \frac{x}{2} \right) = t\]
\[ \Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right) \times \frac{1}{2}dx = dt\]
\[ \Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right)dx = 2dt\]
\[ \therefore I = 2\int \frac{dt}{t^2 - 4t + 1}\]
\[ = 2\int \frac{dt}{t^2 - 4t + 4 - 4 + 1}\]
\[ = 2 \int \frac{dt}{\left( t - 2 \right)^2 - 3}\]

\[ = 2 \int \frac{dt}{\left( t - 2 \right)^2 - \left( \sqrt{3} \right)^2}\]
\[ = 2 \times \frac{1}{2\sqrt{3}}\text{ ln }\left| \frac{t - 2 - \sqrt{3}}{t - 2 + \sqrt{3}} \right| + C\]
\[ = \frac{1}{\sqrt{3}}\text{ ln} \left| \frac{\tan \left( \frac{x}{2} \right) - 2 - \sqrt{3}}{\tan \left( \frac{x}{2} \right) - 2 + \sqrt{3}} \right| + C\]