MCQ
$\int_0^{\pi /4} {[\sqrt {\tan x} + \sqrt {\cot x} ]\,dx} =$
- A$\sqrt 2 \pi $
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{{\sqrt 2 }}$
- D$2\pi $
$ = \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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