Question

\[\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]

Answer 1

\[\text{ Let I } = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]

\[ = \int e^x \left( \frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right) dx\]

\[ = \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx\]

\[ = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx\]

\[ \text{ Putting e}^x \tan \frac{x}{2} = t\]

\[\text{ Diff  both  sides w . r . t . x }\]

\[ e^x \cdot \tan \left( \frac{x}{2} \right) + e^x \times \frac{1}{2} \sec^2 \frac{x}{2} = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left[ \tan \frac{x}{2} + \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) \right] dx = dt\]

\[ \therefore \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx = \int dt\]

\[ = t + C\]

\[ = e^x \tan\left( \frac{x}{2} \right) + C\]