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