# How do I evaluate int1/[x^2(sqrt(25-x^2))] dx ?

int 1/(x^2 sqrt(25-x^2)) dx =?

IN PARTICULAR (this case)

Note that $25 - {x}^{2} \ge 0 \iff x \in \left[- 5 , 5\right]$.

In this case, the useful substitution is

$5 \sin \theta = x$

and this is invertible, because for $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ the function $x \left(\theta\right) = 5 \sin \theta$ is invertible with inverse function $\theta \left(x\right) = \arcsin \left(\frac{x}{5}\right)$, where $x \in \left[- 5 , 5\right]$. So this is a valid substitution.

Let's compute the differential term

$5 \cos \theta \cdot d \theta = \mathrm{dx}$

and turn the integral to $\theta$, substituting the expressions found:

$\int \frac{1}{{x}^{2} \sqrt{25 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{25 {\sin}^{2} \theta \sqrt{25 - 25 {\sin}^{2} \theta}} 5 \cos \theta \cdot d \theta = \int \frac{\cos \theta \cdot d \theta}{5 {\sin}^{2} \theta \cdot 5 \sqrt{1 - {\sin}^{2} \theta}} = \frac{1}{25} \int \frac{\cos \theta \cdot d \theta}{{\sin}^{2} \theta \cos \theta} = - \frac{1}{25} \int - \frac{d \theta}{{\sin}^{2} \theta} = - \frac{1}{25} \cot \theta + C$

Now back to $x$:

$\int \frac{1}{{x}^{2} \sqrt{25 - {x}^{2}}} \mathrm{dx} = - \frac{1}{25} \cot \theta + C = - \frac{1}{25} \cot \left(\arcsin \left(\frac{x}{5}\right)\right) + C$

From trigonometry $\cot \left(\arcsin \left(\frac{x}{5}\right)\right) = \frac{\sqrt{1 - {\left(\frac{x}{5}\right)}^{2}}}{\frac{x}{5}}$.
In fact, you can think of a right triangle with hypotenuse of length $1$ and one leg of length $\frac{x}{5}$ for $0 < x < 5$. By Pythagorean Theorem the length of the other leg is $\sqrt{1 - {\left(\frac{x}{5}\right)}^{2}}$. If we call $\alpha$ the angle between this $\sqrt{1 - {\left(\frac{x}{5}\right)}^{2}}$ leg and the hypotenuse, where $0 < \alpha < \frac{\pi}{2}$, then by definition of sine we have that $\sin \alpha = \frac{\frac{x}{5}}{1} = \frac{x}{5}$, which means that $\alpha = \arcsin \left(\frac{x}{5}\right)$. Moreover, $\cot \left(\alpha\right) = \frac{\sqrt{1 - {\left(\frac{x}{5}\right)}^{2}}}{\frac{x}{5}}$.
So if we substitute $\alpha$ in this equality we get the result for $0 < x < 5$. The result can be extended to $x \in \left[- 5 , 5\right]$ thanks to the symmetry of the functions involved or by re-thinking this argument on the unit circle.

In the end, the result is:
$\int \frac{1}{{x}^{2} \sqrt{25 - {x}^{2}}} \mathrm{dx} = - \frac{1}{25} \frac{\sqrt{1 - {\left(\frac{x}{5}\right)}^{2}}}{\frac{x}{5}} + C = - \frac{1}{25} \frac{\sqrt{25 - {x}^{2}}}{x} + C$

IN GENERAL

When facing expressions like $\sqrt{a - b {x}^{2}}$, where $a , b > 0$, you can take advantage of trigonometric substitutions. Remember that the existence of the radical is determined by the non-negativity of $a - b {x}^{2}$. We have that
$a - b {x}^{2} \ge 0 \iff - \sqrt{\frac{a}{b}} \le x \le \sqrt{\frac{a}{b}}$
This will be important to guarantee that the substitution works with invertible functions.

Let's work on the radical and rewrite it in the following way:
$\sqrt{a - b {x}^{2}} = \sqrt{a \left[1 - \frac{b}{a} {x}^{2}\right]} = \sqrt{a \left[1 - {\left(\sqrt{\frac{b}{a}} x\right)}^{2}\right]} = \sqrt{a} \sqrt{1 - {\left(\sqrt{\frac{b}{a}} x\right)}^{2}}$

Now, if $\sin \theta = \sqrt{\frac{b}{a}} x$, we can take advantage of the Pythagorean Trigonometric Identity
${\cos}^{2} \theta + {\sin}^{2} \theta = 1 \iff {\cos}^{2} \theta = 1 - {\sin}^{2} \theta = 1 - {\left(\sqrt{\frac{b}{a}} x\right)}^{2}$
This means that
$\sqrt{a - b {x}^{2}} = \sqrt{a} \sqrt{1 - {\left(\sqrt{\frac{b}{a}} x\right)}^{2}} = \sqrt{a} \sqrt{{\cos}^{2} \theta}$

Notice that for $\theta \setminus \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ the function $\sin \theta$ is invertible and $\sin \theta \in \left[- 1 , 1\right]$. Moreover, for $x \in \left[- \sqrt{\frac{a}{b}} , \sqrt{\frac{a}{b}}\right]$, the function $\sqrt{\frac{b}{a}} x$ is invertible and $\sqrt{\frac{b}{a}} x \in \left[- 1 , 1\right]$. So the substitution holds, because we're working with invertible functions mapped to the same interval.

The cosine function is non-negative for $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, so we get that

$\sqrt{a - b {x}^{2}} = \sqrt{a} \cos \theta$

So the substitution $\sin \theta = \sqrt{\frac{b}{a}} x$ always "removes" radicals of this kind.