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Q.
The value of the definite integral $\int\limits_0^{\pi / 2} \sqrt{\tan x} d x$, is
Integrals
Solution:
$I =\int\limits_0^{\pi / 2} \sqrt{\tan x} d x \ldots(1) ; I =\int\limits_0^{\pi / 2} \sqrt{\cot x} d x$
adding (1) and (2), we get
$2 I=\int\limits_0^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\sqrt{2} \int\limits_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x=\sqrt{2} \int\limits_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} d x$
$=\sqrt{2} \int\limits_{-1}^1 \frac{ dt }{\sqrt{1- t ^2}}=2 \sqrt{2} \int\limits_0^1 \frac{ dt }{\sqrt{1- t ^2}}=\sqrt{2} \pi ($ where $\sin x -\cos x = t )$
$\therefore I =\frac{\pi}{\sqrt{2}}$