Question

$\int \frac{1}{1+2\cos x}dx$

Answer 1

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