Question

$\int \frac{1}{5-4\cos x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{5-4\cos x}dx \\ \mathrm{Putting}\cos x=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \Rightarrow I=\int \frac{1}{5-4{\displaystyle \left(\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}\right)}}dx \\ =\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}{5\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)-4+4{\tan }^2{\displaystyle \frac{x}{2}}}dx \\ =\int \frac{{\sec }^2\left({\displaystyle \frac{x}{2}}\right)}{9{\tan }^2{\displaystyle \frac{x}{2}}+1}dx \\ \mathrm{Let}\tan \left(\frac{x}{2}\right)=t \\ \Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore I=2\int \frac{dt}{9t^2+1} \\ =\frac{2}{9}\int \frac{dt}{t^2+{\displaystyle \frac{1}{9}}} \\ =\frac{2}{9}\int \frac{dt}{t^2+{\left({\displaystyle \frac{1}{3}}\right)}^2} \\ =\frac{2}{9}\times 3{\tan }^{-1}\left(\frac{t}{{\displaystyle \frac{1}{3}}}\right)+C \\ =\frac{2}{3}{\tan }^{-1}\left(3t\right)+C \\ =\frac{2}{3}{\tan }^{-1}\left(3\tan \frac{x}{2}\right)+C \end{gathered}$$