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

Feb 1, 2017

The integral equals $- \frac{\sqrt{4 - {x}^{2}}}{x} - \arcsin \left(\frac{x}{2}\right) + C$

#### Explanation:

Use trigonometric substitution. We have an integral with a square root of the form $\sqrt{{a}^{2} - {x}^{2}}$. Therefore, use the substitution $x = a \sin \theta$.

In our case, $a = 2$. Our substitution will therefore be $x = 2 \sin \theta$. Then $\mathrm{dx} - 2 \cos \theta d \theta$.

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

$\implies \int \frac{\sqrt{4 - 4 {\sin}^{2} \theta}}{4 {\sin}^{2} \theta} \cdot 2 \cos \theta d \theta$

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

Use ${\sin}^{2} x + {\cos}^{2} x = 1$:

$\implies \int \frac{\sqrt{4 {\cos}^{2} \theta}}{4 {\sin}^{2} \theta} \cdot 2 \cos \theta d \theta$

$\implies \int \frac{2 \cos \theta}{4 {\sin}^{2} \theta} \cdot 2 \cos \theta d \theta$

$\implies \int \frac{4 {\cos}^{2} \theta}{4 {\sin}^{2} \theta} d \theta$

$\implies \int {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta d \theta$

Use $\cot x = \frac{1}{\tan} x = \frac{1}{\sin \frac{x}{\cos} x} = \cos \frac{x}{\sin} x$:

$\implies \int {\cot}^{2} \theta d \theta$

Use ${\cot}^{2} x + 1 = {\csc}^{2} x$:

$\implies \int {\csc}^{2} \theta - 1 d \theta$

$\implies \int {\csc}^{2} \theta d \theta - \int 1 d \theta$

These are both trivial integrals.

$\implies - \cot \theta - \theta + C$

We know from our original substitution that $\frac{x}{2} = \sin \theta$. We can deduce that $\theta = \arcsin \left(\frac{x}{2}\right)$. Therefore, the side adjacent $\theta$ is $\sqrt{4 - {x}^{2}}$ in measure. Thus, $\cot \theta = \frac{\sqrt{4 - {x}^{2}}}{x}$.

$\implies - \frac{\sqrt{4 - {x}^{2}}}{x} - \arcsin \left(\frac{x}{2}\right) + C$

Hopefully this helps!