Help to solve: $\int 1/(x\sqrt{25-x^2})\ dx$ I'm a brand new student. Need some help to integrate this.

Perform this integration:  $$\int \frac{1}{x\sqrt{25-x^2}}\ dx$$

I'm able to obtain in theta terms like this:
$$\frac 15  \ln⁡|\cscθ-\cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:

Using the substitution $x=5\sin{\theta}$ and $dx=5\cos{\theta}$, we obtain:
  $$\int \frac{1}{5\sin{\theta}(5\cos{\theta})}\cdot 5\cos{\theta}~d\theta$$
  Simplifying, we obtain:
  $$\frac{1}{5}\int \frac{d\theta}{\sin{\theta}}$$
  Using trigonometric identities:
  $$\frac{1}{5}\int \csc{\theta}~d\theta$$
  Integrating:
  $$\frac{1}{5}\ln|\csc{\theta}-\cot{\theta}|+C$$


HINT:
If trigonometric substitution is not mandatory,  set $\sqrt{a^2-x^2}=y$
$$\int\dfrac{2x\ dx}{2x^2\sqrt{a^2-x^2}}=\int\dfrac{dy}{y^2-a^2}=?$$
Else setting $x=5\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$\cos y=\sqrt{1-\left(\dfrac x5\right)^2}=\dfrac{\sqrt{25-x^2}}5$
$$\implies\int\dfrac{dx}{x\sqrt{25-x^2}}=\int\dfrac{5\cos y\ dy}{5\sin y(5\cos y)}=\dfrac{\ln|\csc y-\cot y|}5+K$$
