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.
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 $x$.