Question

Evaluate : ${\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}+\sqrt{\cot x})dx=?$

• A. $\frac{\pi }{\sqrt{2}}$
• B. $\frac{\frac{\pi }{2}}{\sqrt{2}}$
• C. $\frac{\pi }{4}\sqrt{2}$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{\sqrt{2}}$

$I={\int }_0^{\pi /4}(\sqrt{tanx}+\sqrt{cotx})dx={\int }_0^{\pi /4}\frac{sinx+cosx}{\sqrt{sinxcosx}}dx=\sqrt{2}{\int }_0^{\pi /4}\frac{sinx+cosx}{\sqrt{sin2x+1-1}}dx=\sqrt{2}{\int }_0^{\pi /4}\frac{sinx+cosx}{\sqrt{1-(sinx-cosx{)}^2}}dx$

Put $sinx-cosx=t\Rightarrow (cosx+sinx)dx=dt$

$I=\sqrt{2}{\int }_{-1}^0\frac{dt}{\sqrt{1-t^2}}=\sqrt{2}{si{n}^{-1}t}_{-1}^0=\sqrt{2}0+\pi /2I=\pi /\sqrt{2}$