# How do you find b such that intdx/(1+x^2)=pi/3 from [0,b]?

Oct 30, 2016

$b = \sqrt{3}$

#### Explanation:

You need to use the standard result $\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan x + C$

We have ${\int}_{0}^{b} \frac{1}{1 + {x}^{2}} \mathrm{dx} = \frac{\pi}{3}$

$\therefore {\left[\arctan x\right]}_{0}^{b} = \frac{\pi}{3}$
$\therefore \arctan b - \arctan 0 = \frac{\pi}{3}$
$\therefore \arctan b - 0 = \frac{\pi}{3}$
$\therefore \arctan b = \frac{\pi}{3}$
$\therefore b = \tan \left(\frac{\pi}{3}\right)$
$\therefore b = \sqrt{3}$