Question

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

Answer 1

\[\text{ Let I } = \int\frac{1}{5 - 4 \sin x}dx\]

\[\text{ Putting sin x} = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]

\[ \therefore I = \int\frac{1}{5 - 4 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]

\[ = \int\frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{5 \left( 1 + \tan^2 \frac{x}{2} \right) - 8 \tan \frac{x}{2}}dx\]

\[ = \int\frac{\sec^2 \frac{x}{2}}{5 \tan^2 \frac{x}{2} - 8 \tan \frac{x}{2} + 5}dx\]

\[\text{ Putting  tan }\frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \text{ sec}^2 \left( \frac{x}{2} \right) dx = dt\]

\[\Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right) dx = \text{ 2 dt}\]
\[ \therefore I = 2\int\frac{1}{5 t^2 - 8t + 5}dt\]
\[ = \frac{2}{5}\int\frac{1}{t^2 - \frac{8}{5}t + 1}dt\]
\[ = \frac{2}{5}\int\frac{1}{t^2 - \frac{8t}{5} + \left( \frac{4}{5} \right)^2 - \left( \frac{4}{5} \right)^2 + 1}dt\]
\[ = \frac{2}{5}\int\frac{1}{\left( t - \frac{4}{5} \right)^2 - \frac{16}{25} + 1}dt\]
\[ = \frac{2}{5}\int\frac{1}{\left( t - \frac{4}{5} \right)^2 + \left( \frac{3}{5} \right)^2}dt\]
\[ = \frac{2}{5} \times \frac{5}{3} \text{ tan}^{- 1} \left( \frac{t - \frac{4}{5}}{\frac{3}{5}} \right) + C .................\left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \text{ tan}^{- 1} \frac{x}{a} + C \right]\]
\[ = \frac{2}{3} \text{ tan}^{- 1} \left( \frac{5t - 4}{3} \right) + C\]
\[ = \frac{2}{3} \text{ tan}^{- 1} \left( \frac{5 \tan \frac{x}{2} - 4}{3} \right) + C ......\left[ \because t = \tan \frac{x}{2} \right]\]