Question

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]

Answer 1

\[\text{We have}, \]

\[I = \int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- \frac{x}{2}} \text{ dx }\]

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

\[ = \int\frac{\sqrt{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2} e^{- \frac{x}{2}}}{2 \cos^2 \frac{x}{2}} \text{ dx }\]

\[ = \int\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{2 \cos^2 \frac{x}{2}} e^{- \frac{x}{2}} \text{ dx }\]

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

\[\text{ Let e}^{- \frac{x}{2}} \sec \left( \frac{x}{2} \right) = t\]

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

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

\[ \therefore I = - \int dt\]

\[ = - t + C\]

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