Question

\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]

Answer 1

\[\text{ Let I }= \int \frac{1}{5 + 7 \cos x + \sin x} \text{ dx }\]
\[\text{ Putting cos x } = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and  sin x }= \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow I = \int \frac{1}{5 + 7 \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \int \frac{\sec^2 \left( \frac{x}{2} \right)}{5\left( 1 + \tan^2 \frac{x}{2} \right) + 7 - 7 \tan^2 \frac{x}{2} + 2 \tan \left( \frac{x}{2} \right)}dx\]
\[ = \int \frac{\sec^2 \left( \frac{x}{2} \right)}{- 2 \tan^2 \left( \frac{x}{2} \right) + 2 \tan \left( \frac{x}{2} \right) + 12}dx\]
\[\text{ Let tan }\left( \frac{x}{2} \right) = t\]
\[ \Rightarrow \frac{1}{2} \text{ sec}^2 \left( \frac{x}{2} \right)dx = dt\]
\[ \Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right)dx = 2dt\]
 `therefore I =∫   {2    dt}/{- 2 t^2 + 2t + 12} `
\[ = \int \frac{dt}{- t^2 + t + 6}\]
\[ = \int \frac{- dt}{t^2 - t - 6}\]
\[ = \int \frac{- dt}{t^2 - t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 6}\]
\[ = \int \frac{- dt}{\left( t - \frac{1}{2} \right)^2 - \frac{1}{4} - 6}\]
\[ = \int \frac{- dt}{\left( t - \frac{1}{2} \right)^2 - \left( \frac{5}{2} \right)^2}\]
\[ = \int \frac{dt}{\left( \frac{5}{2} \right)^2 - \left( t - \frac{1}{2} \right)^2}\]
\[ = \frac{1}{2 \times \frac{5}{2}}\text{ log }\left| \frac{\frac{5}{2} + t - \frac{1}{2}}{\frac{5}{2} - t + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{5}\text{ log }\left| \frac{2 + t}{3 - t} \right| + C\]
\[ = \frac{1}{5}\text{ log } \left| \frac{2 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C\]