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Choose the correct answer in $\int\limits_1^\sqrt 3\frac{dx}{1+x^2}\;equals$\[(A)\;\frac{\pi}{3}\qquad(B)\;\frac{2\pi}{3}\qquad(C)\;\frac{\pi}{6}\qquad(D)\;\frac{\pi}{12}\]

1 Answer

Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
  • (ii)$\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1} (x)+c$
$\int\limits_1^\sqrt 3\large\frac{dx}{1+x^2}$
 
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1} (x)$
 
Where a=1
 
On integrating,
 
Therefore $I=[\tan^{-1}(x)]_1^{v3}$
 
On applying limits,
 
$\tan^{-1}(\sqrt 3)-\tan^{-1}(1)$
 
But $ \tan^{-1}(\sqrt 3)=\frac{\pi}{3}$ and $\tan^{-1}(1)=\frac{\pi}{4}$
 
Therefore $ \large\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$
 
Hence the correct answer is D

 

answered Feb 12, 2013 by meena.p
 
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