Question

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

Answer 1

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