Find $\int_{-5}^{-3} \frac{dx}{\sqrt{x^2 - 4}}$ using trig substitution

Question

$$\int_{-5}^{-3} \frac{dx}{\sqrt{x^2 - 4}}\\ x = 2\sec\theta \\ dx = 2\sec\theta \tan\theta \,d\theta\\ \theta \in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]\\ $$

$$\int_{sec^{-1}\frac{-5}{2}}^{\sec^{-1}\frac{-3}{2}} \frac{2\sec\theta \tan\theta \,d\theta}{\sqrt{4\sec^2\theta - 4}} = \int_{\sec^{-1}\frac{-5}{2}}^{\sec^{-1}\frac{-3}{2}} \frac{\sec\theta \tan\theta\, d\theta}{\sqrt{\sec^2\theta - 1}} = \int_{\sec^{-1}\frac{-5}{2}}^{\sec^{-1}\frac{-3}{2}} \frac{\sec\theta \tan\theta d\theta}{\sqrt{\tan^2\theta}}\\ = \int_{\sec^{-1}\frac{-5}{2}}^{\sec^{-1}\frac{-3}{2}} \frac{\sec\theta \tan\theta d\theta}{|\tan\theta|}$$

From there I don't know if $|\tan\theta| = \tan\theta\,$ or $-\tan\theta$ since $\theta$ can be in the first or second quadrant

Answer 1

It's much simpler with a hyperbolic substitution:

Set $x=2\cosh t,\; t\ge 0\iff t=\operatorname{argcosh}\dfrac x2=\ln\Bigl(\dfrac{x+\sqrt{x^2-4}}2\Bigr),\;x\ge 1$. We get $\mathrm d\mkern1mux=2\sinh t\,\mathrm d\mkern1mu t$, so that $$\int\frac{\mathrm d\mkern1mux}{\sqrt{x^2 - 4}}=\int\frac{2\sinh t\,\mathrm d\mkern1mu t}{2\sqrt{\cosh^2t - 1}}=\int\mathrm d\mkern1mu t=\ln\Bigl(\dfrac{x+\sqrt{x^2-4}}2\Bigr)$$ and finally $$\int_{-5}^{-3} \frac{dx}{\sqrt{x^2 - 4}}=\ln\biggl(\frac{\sqrt{21}-5}{\sqrt{5}-3}\biggr).$$

Answer 2

Let $x=2cosh^{-1}(u/2)$ and then continue

Answer 3

Look at the function $arcsec(x)$ between -5/2 and -3/2 (for example here). It grows and is contained between $\pi/2$ and $\pi$.

This should help you find the quadrants in which $\theta$ really is.

Answer 4

Expanding on my earlier comment to @Bernard, in the interval of integration $x<0$, so we should use $x=-2\cosh t$. Then if we are careful about inverse hyperbolic functions, $$\frac{e^t+e^{-t}}2=-\frac x2$$ $$e^{2t}+xe^t+1=0$$ $$e^t=\frac{-x\pm\sqrt{x^2-4}}2$$ We could choose either sign at this point, but if we choose the $+$ sign, then $e^t>1$ so $t>0$ so $\sinh t>0$ and $$\sqrt{x^2-4}=\color{red}{+}2\sinh t$$ Then $$\int\frac{dx}{\sqrt{x^2-4}}=\int\frac{-2\sinh t\,dt}{2\sinh t}=\int-dt=-t+C=-\ln\left(\frac{-x+\sqrt{x^2-4}}2\right)+C$$ So $$\begin{align}\int_{-5}^{-3}\frac{dx}{\sqrt{x^2-4}}&=\left.-\ln\left(\frac{-x+\sqrt{x^2-4}}2\right)\right|_{-5}^{-3}\\ &=-\ln\left(\frac{3+\sqrt{5}}2\right)+\ln\left(\frac{5+\sqrt{21}}2\right)\\ &=\ln\left(\frac{5+\sqrt{21}}{3+\sqrt5}\right)\end{align}$$ My answer is the additive inverse of @Bernard's answer, but our sign is correct because we have been more careful about domains throughout.