_ ^ x a^3 - x^3 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt {\frac{x}{{{a^3} - {x^3}}}} \;dx = } $

A
${\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{3/2}} + c$
✓
$\frac{2}{3}{\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{3/2}} + c$
C
$\frac{3}{2}{\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{3/2}} + c$
D
$\frac{3}{2}{\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{2/3}} + c$

✓

Answer

Correct option: B.

$\frac{2}{3}{\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{3/2}} + c$

b
(b)Put $x = a{(\sin \theta )^{2/3}} \Rightarrow dx = \frac{2}{3}a{(\sin \theta )^{ - 1/3}}\cos \theta \,d\theta $
$\int_{}^{} {\sqrt {\frac{x}{{{a^3} - {x^3}}}} \,dx} = \int_{}^{} {\frac{{{a^{1/2}}{{(\sin \theta )}^{1/3}}\frac{2}{3}a\,{{(\sin \theta )}^{ - 1/3}}\cos \theta }}{{\sqrt {{a^3} - {a^3}{{\sin }^2}\theta } }}} \,d\theta $
$ = \frac{2}{3}{a^{3/2}}\int_{}^{} {\frac{{\cos \theta \,d\theta }}{{{a^{3/2}}\sqrt {1 - {{\sin }^2}\theta } }}} = \frac{2}{3}{\sin ^{ - 1}}{\left( {\frac{x}{a}} \right)^{3/2}} + c$.

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