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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int_1^{\sqrt 3} \large\frac{dx}{1+x^2} $

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Toolbox:
  • $\int \large\frac{dx}{1+x^2}$$=\tan^{-1}x$
Step 1:
Given :
$\int \large\frac{dx}{1+x^2}$
On integrating we get,
$\big[\tan^{-1}x\big]_1^{\sqrt 3}$
Step 2:
On applying limits we get,
$\tan^{-1}\sqrt 3=\tan^{-1}1$
But $\tan^{-1}\sqrt 3=\large\frac{\pi}{3}$
$\tan^{-1}1=\large\frac{\pi}{4}$
Hence $I=\large\frac{\pi}{3}-\frac{\pi}{4}$
$\Rightarrow \large\frac{\pi}{12}$
answered Nov 11, 2013 by sreemathi.v
 
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