Question

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

Answer 1

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