Question

Assertion : $I={\int }_0^{\frac{\pi }{2}}\sqrt{\tan x}dx=\frac{\pi }{\sqrt{2}}$
Reason: $\tan x=t^2$ makes the integrand in $I$ as a rational function.

• A. Assertion is correct, Reason is correct; Reason is a correct explanation for assertion
• B. Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
• C. Assertion is correct, Reason is incorrect
• D. Assertion is incorrect, Reason is correct

Answer 1

Answer: (A)

$<br/>I={\int }_0^{\frac{\pi }{2}}2\sqrt{\tan x}dx,\text{ Put }\tan x=t^2\Rightarrow dx=\frac{2tdt}{1+t^4}<br/>$
If $x=0\Rightarrow t=0$ and $x=\frac{\pi }{2}\Rightarrow t=\infty$
$<br/>\begin{array}{l}<br/>I={\int }_0^{\infty }\frac{2t^{2}dt}{1+t^4}={\int }_0^{\infty }\frac{t^2+1+t^2-1}{1+t^4}dt\\ <br/>={\int }_0^{\infty }\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt+{\int }_0^{\infty }\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt\\ <br/>={\int }_0^{\infty }\frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)+2}+{\int }_0^{\infty }\frac{d\left(t+\frac{1}{t}\right)}{{\left(t+\frac{1}{t}\right)}^2-2}dt\\ <br/>={\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)|}_0^{\infty }+\frac{1}{2\sqrt{2}}\mathrm{In}{\left(\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)}_0^{\infty }\\ <br/>=\frac{\pi }{\sqrt{2}}<br/>\end{array}<br/>$