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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $ \int \large\frac{8}{9} \frac{x}{\sqrt {1-x^4}}$$dx$

\[\begin {array} {1 1} (a)\;\frac{4}{9} \cos ^{-1} (x^2) +c \\ (b)\;\frac{4}{9} \sin ^{-1} (x^2) +c \\ (c)\;\frac{4}{9} \tan^{-1} (x^2) +c \\ (d)\;None \end {array}\]
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1 Answer

$x^2=t=>2x dx =dt$
$\int \large\frac{4}{9}. \frac{dt}{\sqrt {1-t^2}}$
$ \large\frac{4}{9}.$$\int \large\frac{1}{\sqrt {1-t^2}}$$dt$
$ \large\frac{4}{9}.$$\sin ^{-1} \bigg(\large\frac{t}{1}\bigg) $$+c$
$ \large\frac{4}{9}.$$\sin ^{-1}(x^2)+c$
Hence b is the correct answer.
answered Dec 19, 2013 by meena.p
 
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