Integrating with trig substitution I am trying to integrate $$\int\frac{\sqrt{4-x^2}}{x^2}\ dx$$ I am using the substitution $x=2\sin(\theta)$ and $dx=2\cos(\theta)\ d\theta$. I am a bit confused. If I put the $2$ on the $dx$ side,  I'll get: $\frac12\int\frac{\sqrt{4-4\sin^2\theta}}{4\sin^2\theta}\ \cos\theta\ d\theta$. If I keep the $2$ where it is, I'll get: $\int\frac{\sqrt{4-4\sin^2\theta}}{4\sin^2\theta}\ 2\cos\theta\ d\theta$. These are very different!   

It is the second integral which is correct: you have to express the integrand in function of $\theta$, and $\mathrm dx$ in function of $\mathrm d\theta$.  So you obtain 
2\cos$$\frac{2\sqrt{1-\sin^2\theta}}{4\sin^2\theta}\,2\cos\theta\,\mathrm d\theta$$
Furthermore, the substitution has to be bijective, so we add the condition
$$-\frac\pi 2\le \theta\le \frac\pi 2,\quad(i.e. \theta=\arcsin x),$$
and on this interval, we have $\cos\theta\ge 0$, so the  integral is ultimately, simplifying the coefficients:
$$\int\frac{|\cos\theta|\cos\theta}{\sin^2\theta}\,\mathrm d\theta=\int\frac{\mathrm d\theta}{\tan^2\theta}.$$
Can you take it from here?
