Integral of $$$\frac{1}{x^{2} \sqrt{1 - x^{2}}}$$$

Find $$$\int \frac{1}{x^{2} \sqrt{1 - x^{2}}}\, dx$$$.

Let $$$x=\sin{\left(u \right)}$$$.

Then $$$dx=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(x \right)}$$$.

Therefore,

$$$\frac{1}{x^{2} \sqrt{1 - x^{2}}} = \frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}}$$$

Use the identity $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}}=\frac{1}{\sqrt{\cos^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}}$$$

Assuming that $$$\cos{\left( u \right)} \ge 0$$$, we obtain the following:

$$$\frac{1}{\sqrt{\cos^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}} = \frac{1}{\sin^{2}{\left( u \right)} \cos{\left( u \right)}}$$$

Therefore,

$${\color{red}{\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x}}} = {\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}}$$

Rewrite the integrand in terms of the cosecant:

$${\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}} = {\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}$$

The integral of $$$\csc^{2}{\left(u \right)}$$$ is $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

$${\color{red}{\int{\csc^{2}{\left(u \right)} d u}}} = {\color{red}{\left(- \cot{\left(u \right)}\right)}}$$

Recall that $$$u=\operatorname{asin}{\left(x \right)}$$$:

$$- \cot{\left({\color{red}{u}} \right)} = - \cot{\left({\color{red}{\operatorname{asin}{\left(x \right)}}} \right)}$$

Therefore,

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}$$

Add the constant of integration:

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}+C$$

$$$\int \frac{1}{x^{2} \sqrt{1 - x^{2}}}\, dx = - \frac{\sqrt{1 - x^{2}}}{x} + C$$$A