Evaluate _ ^ x - - x dx - Vidyadip

MCQ

Evaluate $\int\limits_\alpha ^\beta {\sqrt {\frac{{x - \alpha }}{{\beta - x}}} } dx$

A
$\frac{\pi }{2}\left( {\alpha - \beta } \right)$
B
$\frac{\pi }{2}\left( {\alpha + \beta } \right)$
✓
$\frac{\pi }{2}\left( {\beta - \alpha } \right)$
D
$\frac{\pi }{2}\left( {\beta + \alpha } \right)$

✓

Answer

Correct option: C.

$\frac{\pi }{2}\left( {\beta - \alpha } \right)$

c
Let $\mathrm{I}=\int_{\alpha}^{\beta} \sqrt{\frac{\mathrm{x}-\alpha}{\beta-\mathrm{x}}} \mathrm{dx}$

$x=\alpha \cos ^{2} t+\beta \sin ^{2} t$

$\mathrm{x}-\alpha=(\beta-\alpha) \sin ^{2} t$

$\beta-\mathrm{x}=(\beta-\alpha) \cos ^{2} \mathrm{t}$

$(\beta-\alpha) \int_{0}^{\pi / 2}(1-\cos 2 t) d t=(\beta-\alpha)\left[t-\frac{\sin 2 t}{2}\right]_{0}^{\pi / 2}=\frac{\pi}{2}(\beta-\alpha)$

$I=\frac{\pi}{2}(\beta-\alpha)$

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