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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_0^1\frac{dx}{\sqrt{1-x^2}}\]

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Toolbox:
  • $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int \large\frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x+c$
$I= \int \limits_0^1 \int \large \frac{dx}{\sqrt {1-x^2}}$
 
On integrating we get,
 
$\bigg[\sin ^{-1}(x)\bigg]^1_0$
 
On Applying limits we get.
 
$\sin ^{-1}(1)-\sin ^{-1}(0)$
 
But $ \sin^{-1}=\frac{\pi}{2} \qquad \sin ^{-1}(0)=0$
 
Therefore $I=\large\frac{\pi}{2}-0$
 
Hence $ \int \limits_0^1 \large \frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}$

 

answered Feb 11, 2013 by meena.p
 
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