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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate: $\int \large\frac{x^3 \sin ^{-1}(x^2)}{\sqrt {1-x^4}}$$dx$

$(a)\;\frac{-1}{2}\cos^{-1} x^2 (1-x^4)+x^2 +c \\(b)\;\frac{-1}{2}\sin^{-1} x^2 (1-x^4)+x^2 +c \\(c)\;- \sin^{-1} x^2 (1-x^4)+c \\ (d)\;None$
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$\int \large\frac{x^3 \sin ^{-1}(x^2)}{\sqrt {1-x^4}}$$dx$
$\sin ^{-1} (x^2) =t$ => differentiate both sides with respect to x
$\large\frac{1}{\sqrt {1-x^4}} $$ \times 2x dx=dt$
=> $\large\frac{x}{\sqrt {1-x^4}} $$dx=\large\frac{dt}{2}$
=>$ \large\frac{1}{2}$$ \int t. \sin t.dt$
=> $ \large\frac{1}{2}$$ t(- \cos t) + \int \cos t dt$
=> $ \large\frac{-1}{2}$$ t \cos t + \sin t +c$
=>$\large\frac{-1}{2}$$ \sin^{-1} x^2 (1-x^4)+x^2 +c$
Hence b is the correct answer
answered Jan 2, 2014 by meena.p
 
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