_ ^ a - x x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt {\frac{{a - x}}{x}} \;dx = } $

✓
$a\left[ {{{\sin }^{ - 1}}\sqrt {\frac{x}{a}} + \sqrt {\frac{x}{a}} \sqrt {\frac{{a - x}}{a}} } \right] + c$
B
${\sin ^{ - 1}}\frac{x}{a} + \frac{x}{a}\sqrt {{a^2} - {x^2}} + c$
C
$a\left[ {{{\sin }^{ - 1}}\frac{x}{a} - \frac{x}{a}\sqrt {{a^2} - {x^2}} } \right] + c$
D
${\sin ^{ - 1}}\frac{x}{a} - \frac{x}{a}\sqrt {{a^2} - {x^2}} + c$

✓

Answer

Correct option: A.

$a\left[ {{{\sin }^{ - 1}}\sqrt {\frac{x}{a}} + \sqrt {\frac{x}{a}} \sqrt {\frac{{a - x}}{a}} } \right] + c$

a
(a) $I = \int_{}^{} {\sqrt {\frac{{a - x}}{x}} \,dx} $.
Put $x = a{\sin ^2}\theta \Rightarrow dx = 2a\sin \theta \cos \theta \,d\theta ,$ then
$I = \int_{}^{} {\sqrt {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} } \,.\,2a\sin \theta \cos \theta \,d\theta $
$ = a\int_{}^{} {2{{\cos }^2}\theta \,d\theta } = a\int_{}^{} {(1 + \cos 2\theta )\,d\theta } $
$ = a\,\left[ {{{\sin }^{ - 1}}\sqrt {\frac{x}{a}} + \sqrt {\frac{x}{a}} \,.\,\sqrt {\frac{{a - x}}{a}} } \right] + c$.

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