Question

Value of ${\int }_0^{\pi /4}(\sqrt{\tan x}-\sqrt{\cot x})dx$ is

• A. $\sqrt{2}\log (\sqrt{2}-1)$
• B. $\sqrt{2}\log (\sqrt{2}+1)$
• C. $\log (\sqrt{2}+1)$
• D. $\log (\sqrt{2}-1)$

Answer 1

The correct option is A $\sqrt{2}\log (\sqrt{2}-1)$

Let $I={\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}-\sqrt{cotx})dx=-{\int }_0^{\frac{\pi }{4}}\frac{cosx-sinx}{\sqrt{cosxsinx}}dx$
Substitute $\left(sinx+cosx\right)=t\Rightarrow 2sinxcosx=t^2-1$
$\therefore I=\sqrt{2}{\int }_1^{\sqrt{2}}\frac{dt}{\sqrt{t^2-1}}={\left[\sqrt{2}\log (t+\sqrt{t^2}-1)\right]}_0^{\sqrt{2}}=\sqrt{2}\log (\sqrt{2}-1)$