# 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!

• 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. – Jan Jan 5 at 21:40 • When substituting$ x$as$ x=f(t)$you substitute for$dx$with$dx=f'(t)dt\$. So your second integral is the correct substitution. – tmaj Jan 5 at 21:41

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?
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.