# How do you find the integral of int 3/sqrt(1-4x^2)dx?

Apr 1, 2018

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \arcsin \left(2 x\right) + C$

#### Explanation:

Substitute:

$x = \frac{1}{2} \sin t$

$\mathrm{dx} = \frac{1}{2} \cos t \mathrm{dt}$

with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ as the integrand is defined only for $\left\mid 2 x \right\mid < 1$.

Then:

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \int \frac{\cos t \mathrm{dt}}{\sqrt{1 - 4 {\left(\frac{1}{2} \sin t\right)}^{2}}}$

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \int \frac{\cos t \mathrm{dt}}{\sqrt{1 - {\sin}^{2} t}}$

Now, for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$:

$\sqrt{1 - {\sin}^{2} t} = \cos t$

so:

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \int \frac{\cos t \mathrm{dt}}{\cos} t$

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \int \mathrm{dt}$

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3 t}{2} + C$

and undoing the substitution:

$\int \frac{3}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} = \frac{3}{2} \arcsin \left(2 x\right) + C$