_0^ / 4 ( x+ x) d x equals - Vidyadip

MCQ

$\int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x$ equals

A
$\sqrt{2} \pi$
B
$\frac{\pi}{2}$
✓
$\frac{\pi}{\sqrt{2}}$
D
$2 \pi$

✓

Answer

Correct option: C.

$\frac{\pi}{\sqrt{2}}$

(C)
Let $I =\int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x$
$=\int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}} d x$
$=\sqrt{2} \int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} d x$
Put $\sin x-\cos x= t \Rightarrow(\cos x+\sin x) d x= dt$
$\therefore \quad I=\sqrt{2} \int_{-1}^0 \frac{ dt }{\sqrt{1- t ^2}}$
$\begin{array}{l}=\sqrt{2}\left[\sin ^{-1} t\right]_{-1}^0 \\ =\sqrt{2}\left[0-\left(\frac{-\pi}{2}\right)\right]=\frac{\pi}{\sqrt{2}}\end{array}$

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