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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
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Answer

$I=\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }}} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\frac{{\sin x}}{{\cos x}}} }} = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx} } $...(1)
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}dx} $
$\left[ {\because \int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx} } } \right]$
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} $ ...(2)
Adding (1) and (2), we get
$2I = \int_{\pi /6}^{\pi /3} {1dx} $
$= \left[ x \right]_{\pi /6}^{\pi /3}$
$= \frac{\pi }{3} - \frac{\pi }{6} = \frac{\pi }{6}$
$I = \frac{\pi }{{12}}$

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