# How do you integrate int sqrt(4-x^2)/x?

Apr 29, 2018

$- 2 \ln \left\mid \frac{2 + \sqrt{4 - {x}^{2}}}{x} \right\mid - \sqrt{4 - {x}^{2}} + C$

#### Explanation:

$\int \frac{\sqrt{4 - {x}^{2}}}{x} \mathrm{dx}$

Let's use the substitution $x = 2 \sin \theta$. This substitution is motivated by the fact that $\sqrt{4 - {x}^{2}} = \sqrt{4 - 4 {\sin}^{2} \theta} = 2 \sqrt{1 - {\sin}^{2} \theta} = 2 \cos \theta$. This also implies that $\mathrm{dx} = 2 \cos \theta d \theta$.

Using these, the integral becomes:

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

$= 2 \int \left(\csc \theta - \sin \theta\right) d \theta$

These are both well-known integrals:

$= 2 \left(- \ln \left\mid \csc \theta + \cot \theta \right\mid - \cos \theta\right) + C$

We have to return to the variable $x$. Recall that $\sin \theta = \frac{x}{2}$ , so we have a right triangle where the side opposite $\theta$ is $x$, the hypotenuse is $2$, and then by the Pythagorean theorem the side adjacent $\theta$ is $\sqrt{4 - {x}^{2}}$.

Thus $\csc \theta = \frac{2}{x}$, $\cot \theta = \frac{\sqrt{4 - {x}^{2}}}{x}$, and $\cos \theta = \frac{\sqrt{4 - {x}^{2}}}{2}$.

$= - 2 \ln \left\mid \frac{2 + \sqrt{4 - {x}^{2}}}{x} \right\mid - \sqrt{4 - {x}^{2}} + C$

There are a lot of different ways to rewrite the natural logarithm here, but this is fine.