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

Feb 28, 2015

We can evaluate this integral by using a trigonometric substitution.

Let $\sin \theta = 2 x$ Then

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

differentiate

$\mathrm{dx} = \frac{1}{2} \cos \theta d \theta$

Make substitution

$\frac{1}{2} \int \cos \frac{\theta}{\sqrt{1 - 4 {\left(\frac{1}{2} \sin \theta\right)}^{2}}} d \theta$

$\frac{1}{2} \int \cos \frac{\theta}{\sqrt{1 - 4 \left(\frac{1}{4} {\sin}^{2} \theta\right)}} d \theta$

$\frac{1}{2} \int \cos \frac{\theta}{\sqrt{1 - {\sin}^{2} \theta}} d \theta$

$\frac{1}{2} \int \cos \frac{\theta}{\sqrt{{\cos}^{2} \theta}} d \theta$

$\frac{1}{2} \int \cos \frac{\theta}{\cos \theta} d \theta$

$\frac{1}{2} \int d \theta$

Integrating we have $\frac{1}{2} \theta$

Defining $\theta$ in terms of $x$

$\theta = \arcsin \left(2 x\right)$ Therefore

$\frac{1}{2} \arcsin \left(2 x\right) + C$

Feb 28, 2015

The answer is: $\frac{1}{2} \arcsin 2 x + c$.

Remembering that:

$\int \frac{f ' \left(x\right)}{\sqrt{1 - {\left[f \left(x\right)\right]}^{2}}} \mathrm{dx} = \arcsin f \left(x\right) + c$,

so:

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

$= \frac{1}{2} \arcsin 2 x + c$.