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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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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