# How do you integrate int 1/sqrt((1-(x^2)/3))dx using trigonometric substitution?

Sep 7, 2016

$\sqrt{3} \arcsin \left(\frac{x}{\sqrt{3}}\right) + C$

#### Explanation:

We'll use the substitution $x = \sqrt{3} \sin \left(\theta\right)$. Thus, $\mathrm{dx} = \sqrt{3} \cos \left(\theta\right) d \theta$. Substituting into the integral:

$\int \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2} / 3}} = \int \frac{\sqrt{3} \cos \left(\theta\right) d \theta}{\sqrt{1 - \frac{3 {\sin}^{2} \left(\theta\right)}{3}}} = \sqrt{3} \int \frac{\cos \left(\theta\right) d \theta}{\sqrt{1 - {\sin}^{2} \left(\theta\right)}}$

Note that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, so we see that $\cos \left(\theta\right) = \sqrt{1 - {\sin}^{2} \left(\theta\right)}$:

$= \sqrt{3} \int \frac{\cos \left(\theta\right) d \theta}{\cos} \left(\theta\right) = \sqrt{3} \int d \theta = \sqrt{3} \theta + C$

From $x = \sqrt{3} \sin \left(\theta\right)$ we see that $\theta = \arcsin \left(\frac{x}{\sqrt{3}}\right)$:

$= \sqrt{3} \arcsin \left(\frac{x}{\sqrt{3}}\right) + C$