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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!

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    $\begingroup$ If you insert $\mathrm dx /2 = \cos(\theta) \mathrm d \theta$ into the integral, you get $\int \frac{\sqrt{4 - x^2}}{x^2} 2 \frac{\mathrm dx}{2} = \int \frac{\sqrt{4 - \sin^2(\theta)}}{4 \sin^2(\theta)}2 \cos(\theta) \mathrm d \theta$$, which is the same. $\endgroup$ – Jan Jan 5 at 21:40
  • $\begingroup$ When substituting $ x$ as $ x=f(t)$ you substitute for $dx$ with $dx=f'(t)dt$. So your second integral is the correct substitution. $\endgroup$ – tmaj Jan 5 at 21:41
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It is the second integral which is correct: you have to express the integrand in function of $\theta$, and $\mathrm dx$ in function of $\mathrm d\theta$. So you obtain 2\cos$$\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?

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If you put $2$ on the $\mathrm dx$ side, you have $\dfrac{\mathrm dx}2 = \mathrm \cos\theta\,\mathrm d\theta$. Therefore, your integrand must become $$\int\frac{\sqrt{4 - x^2}}{x^2}\cdot2\cdot\frac{\mathrm dx}2$$ for the substitution. You still get $$\int\frac{\sqrt{4 - \sin^2\theta}}{4\sin^2\theta}\cdot2\cos\theta\,\mathrm d\theta$$ in the end.

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