Question

The value of ${\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}+\sqrt{\cot x})dx$ is equal to

• A. $\frac{\pi }{2}$
• B. $-\frac{\pi }{2}$
• C. $\frac{\pi }{\sqrt{2}}$
• D. $-\frac{\pi }{\sqrt{2}}$

Answer 1

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

$\begin{array}{cc}y& ={\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}+\sqrt{\cot x})dx\\ & =\sqrt{2}{\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx\\ & =\sqrt{2}{\int }_0^{\frac{\pi }{4}}\frac{(\sin x+\cos x{)}^{\prime }}{\sqrt{1-(\sin x-\cos x{)}^2}}dx\\ & =\sqrt{2}{\int }_0^{\frac{\pi }{4}}\frac{1}{\sqrt{1-u^2}}du\\ & =\sqrt{2}{\sin }^{-1}u\\ & =\sqrt{2}{\sin }^{-1}(\sin x-\cos x)\end{array}$

Now applying limits, we get

$y=\sqrt{2}{\sin }^{-1}(\sin \frac{\pi }{4}-\cos \frac{\pi }{4})-\sqrt{2}{\sin }^{-1}(\sin 0-\cos 0)=\sqrt{2}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{2}}$

Hence, the answer is $\frac{\pi }{\sqrt{2}}$