Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\sqrt{1+\cos x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{1+\cos x}\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{1+\cos x}\times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1-{\cos }^{2}x}}{\sqrt{1-\cos x}}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{\sqrt{1-\cos x}}dx \\ \mathrm{Let}1-\cos x=u \\ \Rightarrow \sin xdx=du \\ \therefore I=\int \frac{du}{\sqrt{u}} \\ \Rightarrow I=\left[2\sqrt{u}\right] \\ \Rightarrow I={\left[2\sqrt{1-\cos \mathrm{x}}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=2-0 \\ \Rightarrow I=2 \end{gathered}$$