# Trig Substitution Problem - Integration

Suppose I want to integrate this (I chose an easy one): $$\int \frac {dx}{\sqrt{x^2-4}}$$ Method 1: (Trig Substitution) $x=2\sec(\theta)$ $$\int \frac {dx}{\sqrt{x^2-4}}=\int \frac{\sec(\theta)\tan{(\theta)}}{|\tan(\theta)|}d\theta$$

Now removing the absolute value I find is tough. The inverse secant function splits out values of $\theta$ between $0$ and $\pi/2$, and $\pi/2$ to $\pi$. How can I resolve this? Tangent is positive from $0$ and $\pi/2$ - first quadrant, and negative from $\pi/2$ to $\pi$. I know most books overlook this, but I want to be most accurate.

I say "Method 1" because I also have another question right after this.

• Why the absolute value? That is not common place in these types of problems. I suggest dropping the absolute value and cancelling the tangents. This leaves you with just an integral of secant. For that , I highly recommend looking up the answer. That thing is just about impossible to integrate without a ridiculously clever trick. – graydad Sep 2 '14 at 3:09
• Are you trying to find the antiderivative for $x > 2$ or for $x < -2$? – JimmyK4542 Sep 2 '14 at 3:10
• @JimmyK4542, not sure what you mean, but I do want to split the integral from when $\tan(\theta)>0$ and $\tan(\theta)<0$ user166967, we have an absolute value, one does not simply ignore it. Doing that, you are losing half of the solution. I have $|\tan(\theta))|$, which CAN be POSITIVE and NEGATIVE depending on the domain. – CalculusNerd Sep 2 '14 at 3:16
• That being said, I do know the antiderivative of secant, which you can get by multiplying top and bottom by secantx+tangentx. – CalculusNerd Sep 2 '14 at 3:18
• I think you can rewrite this integral by using pythagorean theorem. It eventually became the integral of sec(x). I never encountered a need for absolute value bars. – 123 Sep 2 '14 at 3:20

Notice that the integrand $\dfrac{1}{\sqrt{x^2-4}}$ is only defined on $(-\infty,-2)\cup(2,\infty)$.

Hence, we can consider the antiderivative on $x > 2$ and on $x < -2$ seperately.

If $x > 2$, then we substitute $x = 2\sec \theta$ where $0 < \theta < \tfrac{\pi}{2}$.

Over this range, $\tan \theta > 0$. So, $\sqrt{x^2-4} = 2\tan \theta$, and thus,

$\displaystyle\int\dfrac{dx}{\sqrt{x^2-4}} = \int \dfrac{2\sec \theta \tan \theta}{|2\tan \theta|} \,d\theta = \int \dfrac{\sec \theta \tan \theta}{\tan \theta} \,d\theta = \int \sec \theta \,d\theta$

$= \ln|\sec\theta + \tan \theta|+C = \ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C = \ln\left(\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right)+C$

If $x < -2$, then we substitute $x = 2\sec \theta$ where $\tfrac{\pi}{2} < \theta < \pi$.

Over this range, $\tan \theta < 0$. So, $\sqrt{x^2-4} = -2\tan \theta$, and thus,

$\displaystyle\int\dfrac{dx}{\sqrt{x^2-4}} = \int \dfrac{2\sec \theta \tan \theta}{|2\tan \theta|} \,d\theta = \int \dfrac{\sec \theta \tan \theta}{-\tan \theta} \,d\theta = -\int \sec \theta \,d\theta$

$= -\ln|\sec\theta + \tan \theta|+C = -\ln\left|\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}\right|+C = \ln\left|\dfrac{1}{\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}}\right|+C$.

$= \ln\left|\dfrac{\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}}{\left(\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}\right)\left(\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right)}\right|+C = \ln\left|\dfrac{\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}}{\frac{x^2}{4}-\frac{x^2-4}{4}}\right|+C = \ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C$

As you can see, the two expressions are the same. So, in this case, we don't lose anything by carelessly writing $\sqrt{x^2-4} = 2\tan \theta$ instead of $|2\tan \theta|$.

Alternatively, if $x < -2$, then substitute $x = 2\sec \theta$ where $\pi < \theta < \frac{3\pi}{2}$. Then $\tan \theta > 0$ over this range, and so, we don't need to worry about the absolute value signs.

• Jeez, that was perfect, and the answer I was looking for :D – CalculusNerd Sep 2 '14 at 4:05
• But just a quite question...how does x>2 imply that $0<\theta <\pi/2$? – CalculusNerd Sep 2 '14 at 4:08
• We need to pick a range for $\theta$ such that for every $x > 2$ we can write $x = 2\sec \theta$ for exactly one value of $\theta$ in our range. One such range is $0 < \theta < \pi/2$, although there are several other choices. – JimmyK4542 Sep 2 '14 at 4:12
• I understand that, but I don't see how you figured out that the range must be $0 < \theta < \pi/2$. How did you work that out? – CalculusNerd Sep 2 '14 at 4:18
• Try looking at the graph of $x = 2\sec \theta$. You'll see that as $\theta$ ranges from $0$ to $\pi/2$, $x$ increases from $2$ to $\infty$. – JimmyK4542 Sep 2 '14 at 4:23