Evaluate the following integrals:

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

put $\frac{x}{2}=t \Rightarrow x=2 t \Rightarrow d x=2 d t$

$\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x=2 \int \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} e^{-t} d t$

$=2 \int \frac{\sqrt{\sin ^{2} t+\cos ^{2} t-2 \sin t \cos t}}{1+\cos 2 t} \mathrm{e}^{-t} \mathrm{dt}$

$=2 \int \frac{\sqrt{(\cos t-\sin t)^{2}}}{2 \cos ^{2} t} \mathrm{e}^{-t} \mathrm{dt}$

$=\int(\operatorname{sect}-\tan t \sec t) \mathrm{e}^{-t} \mathrm{dt}$

$=\int \sec t \mathrm{e}^{-t} \mathrm{dt}-\int \tan t \sec t \mathrm{e}^{-t} \mathrm{dt}$

Integrating by parts

$=\mathrm{e}^{-\mathrm{t}} \sec \mathrm{t}+\int \tan \mathrm{t} \sec \mathrm{t} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}-\int \tan \mathrm{t} \sec \mathrm{t} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}$

$=e^{-t} \sec t+c$

$=e^{-\frac{x}{2}} \sec \frac{x}{2}+c$