Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

\[ \Rightarrow I = \int t + C\]

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