Question

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

Answer 1

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