Question

The value of the integral $\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ is
(a) 0
(b) π/2
(c) π/4
(d) none of these

Answer 1

(c) π/4

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}{\sqrt{\cos \left(\frac{\mathrm{\pi }}{2}-x\right)}+\sqrt{\sin \left(\frac{\mathrm{\pi }}{2}-x\right)}}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}}\mathrm{d}\mathrm{x}...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}+\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}}=\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence}I=\frac{\mathrm{\pi }}{4} \end{gathered}$$