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Evaluate : $ \int \large\frac{dx}{\sqrt{1-x^2} } $

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  • $\int\large\frac{dx}{\sqrt{1-x^2}}$$=\sin^{-1}x/a$
$\int\large\frac{dx}{\sqrt{1-x^2}}$$=\sin^{-1}x/a$+c
$I=\sin^{-1}x/a$+c
answered Nov 12, 2013 by sreemathi.v
 
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