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Model Paper 10 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 103 Marks
Question
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
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Answer

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

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