Question

Prove that ${\int }_0^{\pi /4}(\sqrt{\tan x}+\sqrt{\cot x})dx=\sqrt{2}\cdot \frac{\pi }{2}$

Answer 1

Let LHS $={\int }_0^{\frac{\pi }{4}}(\sqrt{tanx}+\sqrt{cotx})dx$

$={\int }_0^{\pi /4}(\sqrt{\tan x}\left(\frac{\sqrt{sinx}}{\sqrt{(}cosx)+\frac{\sqrt{cosx}}{\sqrt{sinx}}}\right)dx={\int }_0^{\pi /4}(\sqrt{\tan x}\frac{sinx+cosx}{\sqrt{sinx.cosx}}$

Thus $\sqrt{2}{\int }_0^{\pi /4}(\sqrt{\tan x}\frac{sinx+cosx}{\sqrt{2sinx.cosx}}dx=\sqrt{2}\int {\int }_0^{\pi /4}(\sqrt{\tan x}\frac{sinx+cosx}{\sqrt{1-(sinx-cosx{)}^2}}$

Let $sinx-cosx=z$

$\Rightarrow (cosx+sinx)dx=dz$

Also If $x=0,z=-1$ and $x=\frac{\pi }{4},z=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0$

Therefore, LHS $=\sqrt{2}{\int }_{-1}^0\frac{dz}{\sqrt{1-z^2}}$

$=\sqrt{2}[si{n}^{-1}z{]}_{-1}^0=\sqrt{2}[si{n}^{-1}0-si{n}^{-1}(-1)]$

$=\sqrt{2}[0-(-\frac{\pi }{2})]=\sqrt{2}.\frac{\pi }{2}$ = RHS.

Hence proved.