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Home  >>  CBSE XII  >>  Math  >>  Integrals

If $\Large \int\frac{1}{1+4x^2}$$dx=\large\frac{\pi}{8},$then a=____________.

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  • $\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_a^0\large\frac{1}{1+4x^2}$$dx$
$\large\frac{1}{1+4x^2}$$dx$ can be written as $\large\frac{1}{4(\Large\frac{1}{4}+\normalsize x^2)}$
$I=\large\frac{1}{4}$$\int_0^a\large\frac{dx}{x^2+\big(\Large\frac{1}{2}\big)^2}$
This is of the form $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\bigg(\Large\frac{x}{a}\big)$$+c.$
Here $x=x$ and $a=\large\frac{1}{2}$
$\Rightarrow=\large\frac{1}{4}[\frac{1}{1/2}$$\tan^{-1}\big(\large\frac{x}{1/2}\big)]_0^a$
$\;\;=\large\frac{1}{4}$$\times 2[\tan^{-1}(2x)]_0^a$
$\;\;=\large\frac{1}{2}$$[\tan^{-1}(2x)]_0^a$
Step 2:
On applying limits,
$\large\frac{1}{2}$$[\tan^{-1}(2a)-\tan^{-1}(0)$]
But $\tan^{-1} 0=0.$
Hence $I=\large\frac{1}{2}$$\tan^{-1}(2a)$
It is given that $\int_0^a\large\frac{1}{1+4x^2}$$dx$=$\large\frac{\pi}{8}$
Now equating I to $\large\frac{\pi}{8}$
$\large\frac{1}{2}$$\tan^{-1}(2a)=\large\frac{\pi}{8}$
Putting a=$\large\frac{1}{2}$
$\Rightarrow \large\frac{1}{2}$$\tan^{-1}(2.\large\frac{1}{2})=\frac{1}{2}$$\tan^{-1}(1)$
But $\tan^{-1}(1)=\large\frac{\pi}{4}$
$\large\frac{1}{2}\frac{\pi}{4}=\frac{\pi}{8}$
Hence the value should be $\large\frac{1}{2}$
answered Apr 25, 2013 by sreemathi.v
 
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