_ ^ a + x a - x ;dx = - Vidyadip

MCQ

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

A
$a = \frac{1}{2}$
B
$a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$
C
$ - a{\cos ^{ - 1}}x/a + \sqrt {{a^2} - {x^2}} + c$
✓
$ - a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$

✓

Answer

Correct option: D.

$ - a{\cos ^{ - 1}}x/a - \sqrt {{a^2} - {x^2}} + c$

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

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