Question

$\int \frac{1}{1-2\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{1-2\sin x}dx \\ \mathrm{Putting}\sin x=\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \Rightarrow I=\int \frac{1}{1-2\times {\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}dx \\ =\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)-4\tan \left({\displaystyle \frac{x}{2}}\right)}dx \\ =\int \frac{{\sec }^2\left({\displaystyle \frac{x}{2}}\right)}{{\tan }^2\left({\displaystyle \frac{x}{2}}\right)-4\tan \left({\displaystyle \frac{x}{2}}\right)+1}dx \\ \mathrm{Let}\tan \left(\frac{x}{2}\right)=t \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)\times \frac{1}{2}dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore I=2\int \frac{dt}{t^2-4t+1} \\ =2\int \frac{dt}{t^2-4t+4-4+1} \\ =2\int \frac{dt}{{\left(t-2\right)}^2-3} \\ =2\int \frac{dt}{{\left(t-2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =2\times \frac{1}{2\sqrt{3}}\ln \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|+C \\ =\frac{1}{\sqrt{3}}\ln \left|\frac{\tan \left({\displaystyle \frac{x}{2}}\right)-2-\sqrt{3}}{\tan \left({\displaystyle \frac{x}{2}}\right)-2+\sqrt{3}}\right|+C \end{gathered}$$