Question

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

Answer 1

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