Question

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

Answer 1

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