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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int_0^1\large\frac{dx}{1+x^2} $

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)Methods of substitution:
  • Let f(x)=t
  • f'(x)dx=dt.
  • $\int f(x)dx=\int x.dt$
  • $\int\large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$$+c$
Step 1:
Let $I=\int _0^1\large\frac{dx}{1+x^2}$
This is of the form
$\int\large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$$+c$
$I=[\tan^{-1}x]_0^1$
Step 2:
On applying limits we get,
$I=\tan^{-1}(1)-\tan^{-1}(0)$
$\;\;=\tan^{-1}(1)$
But $\tan^{-1}(1)=\large\frac{\pi}{4}$
$\therefore I=\large\frac{\pi}{4}$
answered Sep 26, 2013 by sreemathi.v
 
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