Question

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

Answer 1

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