_0^ /4 [ x + x ] ,dx equals - Vidyadip

MCQ

$\int_0^{\pi /4} {[\sqrt {\tan x} + \sqrt {\cot x} ]\,dx} $ 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
(c) $I = \int_0^{\pi /4} {[\sqrt {\tan x} + \sqrt {\cot x]} } dx = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{\sqrt {\sin x\cos x} }}dx} $

$ = \sqrt 2 \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{\sqrt {1 - {{(\sin x - \cos x)}^2}} }}dx} $

Put $\sin x - \cos x = t$; $(\cos x + \sin x)dx = dt$

$\therefore \,\,\,I = \sqrt 2 \int_{ - 1}^0 {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} $

$I = \sqrt 2 [{\sin ^{ - 1}}t]_{ - 1}^0 = \sqrt 2 [0 - ( - \pi /2)] = \frac{\pi }{{\sqrt 2 }}$.

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