MCQ
$\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }} = } $
- ✓$\pi /12$
- B$\pi /2$
- C$\pi /6$
- D$\pi /4$
$ = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;dx} $ ..$(i)$
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;} $ ..$(ii)$
(Since $\int_a^b {f(x)dx} = \int_a^b {f(a + b - x)\,dx} $)
Adding $(i)$ and $(ii),$ we get,
$2I = \int_{\pi /6}^{\pi /3} {\;dx} $
==> $I = \frac{1}{2}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{12}}$.
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$ + sin^{-1} \left\{ \,\frac{1}{{\sqrt {13} }}(2\cos x + 3\sin x)\,\,\,\right\} $ w.r.t. at $x = \frac{3}{4}$ is :
$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
$\frac{\pi}{2}+\log\big(2\sqrt{2}\big)$
$\frac{\pi}{6}+\log\big(2\sqrt{2}\big)$
$\frac{\pi}{3}+\log\big(2\sqrt{2}\big)$