_ ^ ( 1 - x 1 + x ) ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt {\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)} } \;dx = $

✓
${\cos ^{ - 1}}\sqrt x + \sqrt {1 - x} \;.\;(\sqrt x - 2) + c$
B
${\cos ^{ - 1}}\sqrt x - \sqrt {1 - x} \;.\;(\sqrt x - 2) + c$
C
${\cos ^{ - 1}}\sqrt x + \sqrt {1 - x} \;.\;(\sqrt {x - 2} ) + c$
D
None of these

✓

Answer

Correct option: A.

${\cos ^{ - 1}}\sqrt x + \sqrt {1 - x} \;.\;(\sqrt x - 2) + c$

a
(a) Put $x = {\cos ^2}\theta \Rightarrow dx = - 2\cos \theta \sin \theta \,d\theta ,$ then
$\int_{}^{} {\sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \,dx} = - 4\int_{}^{} {{{\sin }^2}\frac{\theta }{2}\cos \theta \,d\theta } $
$ = - 2\int_{}^{} {(1 - \cos \theta )\cos \theta \,d\theta } = \theta + \frac{1}{2}\sin 2\theta - 2\sin \theta $
$ = \theta + \sin \theta \,\cos \theta - 2\sin \theta $
$ = {\cos ^{ - 1}}\sqrt x + (\sqrt {1 - x} )(\sqrt x - 2) + c$

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The correct option is: