Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sin \mathrm{\theta }}{\sqrt{1+\cos \mathrm{\theta }}}d\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \theta }{\sqrt{1+\cos \theta }}\mathit{}\mathit{d}\theta . \\ \mathrm{Let}\cos \theta =t.\mathrm{Then},-\sin \theta d\theta =dt \\ \mathrm{When}\theta =0,t=1\mathrm{and}\theta =\frac{\mathrm{\pi }}{2},t=0 \\ \therefore I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \theta }{\sqrt{1+\cos \theta }}\mathit{d}\theta \\ ={\int }_1^0\frac{-dt}{\sqrt{1+t}} \\ ={\int }_0^1\frac{dt}{\sqrt{1+t}} \\ =2{\left[\sqrt{1+t}\right]}_0^1 \\ =2\left(\sqrt{2}-1\right) \end{gathered}$$