# What is the definite integral of 1/(36+x^2) with bounds [0, 6]?

Jan 10, 2016

$\frac{\pi}{24}$
There is a standard integral: $\int \left(\frac{1}{{a}^{2} + {x}^{2}}\right) \mathrm{dx} = \frac{1}{a} {\tan}^{-} 1 \left(\frac{x}{a}\right) + c$
So ${\int}_{0}^{6} \left(\frac{1}{36 + {x}^{2}}\right) \mathrm{dx} = {\left[\frac{1}{6} {\tan}^{-} 1 \left(\frac{x}{6}\right)\right]}_{0}^{6}$
$= \left(\frac{1}{6} {\tan}^{-} 1 \left(1\right)\right) - \left(\frac{1}{6} {\tan}^{-} 1 \left(0\right)\right) = \frac{1}{6} \cdot \frac{\pi}{4} - \frac{1}{6} \cdot 0 = \frac{\pi}{24}$