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