# How do you integrate ((4-x^2)^(1/2)) / x^2?

Aug 24, 2015

Keeping trigonometric substitution in mind, the numerator is of the form:

$\sqrt{{a}^{2} - {x}^{2}}$

which resembles $\sqrt{1 - {\sin}^{2} x}$. Thus, let:

$x = a \sin \theta$ where $a = 2$
$\implies x = 2 \sin \theta$
${x}^{2} = {a}^{2} {\sin}^{2} \theta = 4 {\sin}^{2} \theta$
$\sqrt{4 - {x}^{2}} = \sqrt{{2}^{2} - {2}^{2} {\sin}^{2} \theta} = 2 \cos \theta$
$\mathrm{dx} = 2 \cos \theta d \theta$

With this substitution, we get:

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

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

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

since $1 + {\cot}^{2} \theta = {\csc}^{2} \theta$, just like how $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$.

The derivative of $\cot \theta$ is $- {\csc}^{2} \theta$, so with some negative-sign manipulation, we can get this into a more simplified form:

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

$= - \left(\int d \theta - \int {\csc}^{2} \theta d \theta\right)$

$= - \left(\int d \theta + \int - {\csc}^{2} \theta d \theta\right)$

$= - \int d \theta - \int - {\csc}^{2} \theta d \theta$

$= - \theta - \cot \theta$

With $x = 2 \sin \theta$ and $\sqrt{4 - {x}^{2}} = 2 \cos \theta$:

$\implies \theta = \arcsin \left(\frac{x}{2}\right)$
$\implies \cot \theta = \cos \frac{\theta}{\sin} \theta = \left(\frac{\sqrt{4 - {x}^{2}}}{\cancel{2}}\right) \left(\frac{\cancel{2}}{x}\right)$
$= \frac{\sqrt{4 - {x}^{2}}}{x}$

So the final answer is:

$= \textcolor{b l u e}{- \arcsin \left(\frac{x}{2}\right) - \frac{\sqrt{4 - {x}^{2}}}{x} + C}$