Question

Assertion : $I=\int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} d x=\frac{\pi}{\sqrt{2}}$
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)

$
I=\int_{0}^{\frac{\pi}{2}} 2 \sqrt{\tan x} d x, \text { Put } \tan x=t^{2} \Rightarrow d x=\frac{2 t d t}{1+t^{4}}
$
If $x=0 \Rightarrow t=0$ and $x=\frac{\pi}{2} \Rightarrow t=\infty$
$
\begin{array}{l}
I=\int_{0}^{\infty} \frac{2 t^{2} d t}{1+t^{4}}=\int_{0}^{\infty} \frac{t^{2}+1+t^{2}-1}{1+t^{4}} d t \\
=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t+\int_{0}^{\infty} \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\
=\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} d t \\
=\left.\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right|_{0} ^{\infty}+\frac{1}{2 \sqrt{2}} \operatorname{In}\left(\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)_{0}^{\infty} \\
=\frac{\pi}{\sqrt{2}}
\end{array}
$