_ ^ 3 x^2 9 - 16 x^6 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{3{x^2}}}{{\sqrt {9 - 16{x^6}} }}} \;dx = $

✓
$\frac{1}{4}{\sin ^{ - 1}}\left( {\frac{{4{x^3}}}{3}} \right) + c$
B
$\frac{1}{3}{\sin ^{ - 1}}\left( {\frac{{4{x^3}}}{3}} \right) + c$
C
$\frac{1}{4}{\sin ^{ - 1}}{x^3} + c$
D
$\frac{1}{3}{\sin ^{ - 1}}{x^3} + c$

✓

Answer

Correct option: A.

$\frac{1}{4}{\sin ^{ - 1}}\left( {\frac{{4{x^3}}}{3}} \right) + c$

a
(a)$\int_{}^{} {\frac{{3{x^2}}}{{\sqrt {9 - 16{x^6}} }}dx = \int_{}^{} {\frac{{3{x^2}}}{{\sqrt {{{(3)}^2} - {{(4{x^3})}^2}} }}\,dx} } $
Put $4{x^3} = t \Rightarrow 12{x^2}dx = dt,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{\sqrt {{{(3)}^2} - {t^2}} }} = \frac{1}{4}.\frac{1}{1}{{\sin }^{ - 1}}\left( {\frac{t}{3}} \right) + c = \frac{1}{4}{{\sin }^{ - 1}}\left( {\frac{{4{x^3}}}{3}} \right)} + c$.

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