Question

\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]

Answer 1

\[\text{  Let I }= \int \frac{1}{13 + 3 \cos x + 4 \sin x}dx\]
\[\text{ Putting cos x }= \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and sin x }= \frac{2\tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}\]
\[ \therefore I = \int \frac{1}{13 + 3 \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + 4 \times 2\frac{\tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}}dx\]
\[ = \int \frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{13\left( 1 + \tan^2 \frac{x}{2} \right) + 3 - 3 \tan^2 \frac{x}{2} + 8 \tan \left( \frac{x}{2} \right)} dx\]
\[ = \int \frac{\sec^2 \frac{x}{2}}{13 \tan^2 \frac{x}{2} - 3 \tan^2 \frac{x}{2} + 16 + 8 \tan \left( \frac{x}{2} \right)}dx\]
\[ = \int \frac{\sec^2 \frac{x}{2}}{10 \tan^2 \left( \frac{x}{2} \right) + 8 \tan \left( \frac{x}{2} \right) + 16}dx\]
\[\text{ Let tan} \left( \frac{x}{2} \right) = t\]
\[ \Rightarrow \frac{1}{2} \text{ sec}^2 \left( \frac{x}{2} \right)dx = dt\]
\[ \Rightarrow \text{ sec}^2 \left( \frac{x}{2} \right)dx = 2dt\]
\[ \therefore I = \int \frac{2 dt}{10 t^2 + 8t + 16}\]
\[ = \int \frac{dt}{5 t^2 + 4t + 8}\]
\[ = \frac{1}{5} \int \frac{dt}{t^2 + \frac{4}{5}t + \frac{8}{5}}\]
\[ = \frac{1}{5}\int \frac{dt}{t^2 + \frac{4}{5}t + \left( \frac{2}{5} \right)^2 - \left( \frac{2}{5} \right)^2 + \frac{8}{5}}\]

\[ = \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 - \frac{4}{25} + \frac{8}{5}}\]
\[ = \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 + \frac{- 4 + 40}{25}}\]
\[ = \frac{1}{5}\int \frac{dt}{\left( t + \frac{2}{5} \right)^2 + \left( \frac{6}{5} \right)^2}\]
\[ = \frac{1}{5} \times \frac{5}{6} \tan^{- 1} \left( \frac{t + \frac{2}{5}}{\frac{6}{5}} \right) + C\]
\[ = \frac{1}{6} \tan^{- 1} \left( \frac{5t + 2}{6} \right) + C\]
\[ = \frac{1}{6} \tan^{- 1} \left( \frac{5 \tan \frac{x}{2} + 2}{6} \right) + C\]