# How do you evaluate int 1/(1+4x^2) from [0,sqrt3/2]?

Nov 24, 2016

You integrate by reconducing to the fundamental integral with arcotangent primitive

#### Explanation:

${\int}_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{dx}}{1 + 4 {x}^{2}} = {\int}_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{dx}}{1 + {\left(2 x\right)}^{2}} = \frac{1}{2} {\int}_{0}^{\sqrt{3}} \frac{\mathrm{dt}}{1 + {t}^{2}} = \frac{1}{2} \arctan \left(t\right) {|}_{0}^{\sqrt{3}} = \frac{1}{2} \left(\arctan \left(\sqrt{3}\right) - \arctan \left(0\right)\right) = \frac{\pi}{6}$

by substituting $t = 2 x$ and therefore $\mathrm{dx} = \frac{\mathrm{dt}}{2}$

(Remember when doing a substitution also the interval changes, so if x is in $\left[0 , \frac{\sqrt{3}}{2}\right]$, t is in $\left[0 , \sqrt{3}\right]$)

Nov 24, 2016

The answer is $= \frac{\pi}{6}$

#### Explanation:

We evaluate this integral by substitution

Let $u = 2 x$

$\mathrm{du} = 2 \mathrm{dx}$

$\int \frac{\mathrm{dx}}{1 + 4 {x}^{2}} = \frac{1}{2} \int \frac{\mathrm{du}}{1 + {u}^{2}}$

Let $u = \tan v$

$1 + {u}^{2} = 1 + {\tan}^{2} v = \frac{1}{\cos} ^ 2 v$

$\mathrm{du} = \frac{\mathrm{dv}}{\cos} ^ 2 v$

$\frac{1}{2} \int \frac{\mathrm{du}}{1 + {u}^{2}} = \frac{1}{2} \int \frac{\mathrm{dv}}{{\cos}^{2} v \cdot \frac{1}{\cos} ^ 2 v}$

$= \frac{1}{2} v = \frac{1}{2} \cdot \arctan u = \frac{1}{2} \arctan 2 x + C$

${\int}_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{dx}}{1 + 4 {x}^{2}} = {\left[\frac{1}{2} \cdot \arctan 2 x\right]}_{0}^{\frac{\sqrt{3}}{2}}$

$= \frac{1}{2} \cdot \arctan \sqrt{3} = \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6}$