Question

$\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\left(\sqrt{\tan }x+\sqrt{\cot }x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right)\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx \\ \Rightarrow I=\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}dx \\ \Rightarrow I=\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin x+\cos x}{\sqrt{1-{\left(\sin x-\cos x\right)}^2}}dx \\ \mathrm{Let}\sin x-\cos x=t.\mathrm{Then},\cos x+\sin xdx=dt \\ \mathrm{When}x=0,t=1\mathrm{and}x=\frac{\mathrm{\pi }}{4},t=0 \\ \therefore I=\sqrt{2}{\int }_{-1}^0\frac{dt}{\sqrt{1-t^2}} \\ \Rightarrow I=\sqrt{2}{\left[{\sin }^{-1}t\right]}_{-1}^0 \\ \Rightarrow I=\frac{\mathrm{\pi }}{\sqrt{2}} \\ \Rightarrow I= \end{gathered}$$