Question

\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]

Answer 1

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