TITLE: Trigonometric substitution for $\int\frac{1}{x^2\sqrt{4-x^2}}dx$
QUESTION [5 upvotes]: I'm reviewing my quizzes to study for midterm tomorrow, and I came across a problem where I'm supposed to integrate:
$$\int\frac{1}{x^2\sqrt{4-x^2}}dx$$
I used Mathematica to solve the problem and I'm sure it gave me the correct answer, which is:
$$-\frac{\sqrt{4-x^2}}{4x}$$
I used $ x = 2\sin{\theta}$ and $dx = 2\cos{\theta}$ $d\theta$ to solve the problem, and I only got to $$\frac{1}{8}\int{\frac{1}{\sin^2{\theta}\cos{\theta}}}d\theta$$
Looking at step-by-step solution via WolframAlpha, they used $\theta = \arcsin{\frac{x}{2}}$ to solve the problem which I do not know how to.
I don't think there is a need for $\theta = \arcsin{\frac{x}{2}}$ to solve the problem, and I'm wondering if anyone can show me how to solve this step by step without the use of $\theta = \arcsin{\frac{x}{2}}$? Maybe help me understand how to?
Trigonometric substitution is the only method that I'm struggling with, and any tips on improving trig sub skill would be appreciated too.
Thanks.

REPLY [2 votes]: Note that if $\theta=\arcsin\frac{x}{2}$ then $x=2\sin\theta$, so they're using the same substitution as you.
When you include your point that $\mathrm{d}x =2\cos\theta\mathrm{d}\theta$, you're looking for $$\frac{1}{4}\int\frac{1}{\sin^2\theta}\mathrm{d}\theta=\frac{-\cos\theta}{4\sin\theta}$$
You should confirm this by differentiating the right hand side.
Now try to draw a triangle which would give rise to the relationship $x=2\sin\theta\,$ in order to obtain an expression in terms of $x$.