# Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$

I am trying to find $$\int \frac {\sqrt {x^2 - 4}}{x} dx$$

I make $x = 2 \sec\theta$

$$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$$

$$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$$

$$\int \frac {2\tan \theta}{x} dx$$

From here I am not too sure what to do but I know I shouldn't have x.

$$\int \frac {2\tan \theta}{2 \sec\theta} dx$$

I also know I shouldn't have dx anymore.

$$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$$

$$\int \frac {2\tan \theta}{2 \sec\theta} 2\sec \theta \tan \theta \; \mathrm d\theta$$

$$\int {2\tan^2 \theta} \; \mathrm d\theta$$

$$2\int {\tan^2 \theta} \; \mathrm d\theta$$

I have no idea how to find the integral of $\tan^2 \theta$

So I use Wolfram Alpha:

$$\tan \theta - \theta + c$$

Now I need to replace theta with x.

$$x = 2 \sec\theta$$

With same mathmagics I produce

$$\frac {x}{2} = \sec \theta$$

$$\theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$$

$$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$$

This is wrong but I am not sure why.

• Congratulations Jordan, you're getting better at this! (But that is not magic, it is maths!) – Pedro Tamaroff Jun 4 '12 at 1:40

You are correct. First note that you have not carried a factor of $2$, since your integral is $2 \int \tan^2(\theta) d \theta$.

Hence your solution should read $$2 \tan(\text{arsec}(x/2)) - 2 \text{arcsec}(x/2) + c$$ You may want to rewrite your solution to match with the solution in your text. For instance, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$

So why is the above identity true? If we let $\text{arsec}(x/2)) = \theta$, then we get that $\sec(\theta) = x/2$ i.e. $\sec^2(\theta) = \dfrac{x^2}{4}$. We have that $\tan^2(\theta) = \sec^2(\theta) - 1 = \dfrac{x^2}{4} - 1 = \dfrac{x^2-4}{4}$ i.e. $\tan(\theta) = \dfrac{\sqrt{x^2 - 4}}{2}$. Hence, $$\theta = \arctan \left( \dfrac{\sqrt{x^2-4}}{2}\right)$$

Hence, we have the identity, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$

If you use this, then your solution will read $$\sqrt{x^2 - 4} - 2 \, \text{arsec}(x/2)) + c = \sqrt{x^2 - 4} - 2 \, \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right) + c$$

you're getting a lot better at these. In fact the answer you have is completely correct, it probably just doesn't look like the answer in the book. You get that with trig functions a lot, since there are often multiple ways of representing the same thing. What answer do you think you should have gotten?

EDIT: Okay. The term that looks different is $tan(arcsec(x/2))$. There are going to be a few heavy trig identities here:

$x=2\sec\theta=\frac 2 {\cos\theta}$, so $arcsec\frac x 2=\cos^{-1}\frac 2 x$. I'm doing this because I'm going to rewrite $\tan$ in terms of $\cos$ and cancel out $\cos(\cos^{-1}\frac 2 x)=\frac 2 x$ $$\tan(\cos^{-1}\frac 2 x)=\frac{\sin(\cos^{-1}\frac 2 x)}{\cos(\cos^{-1}\frac 2 x)}=\frac{\sqrt{1-(\cos(\cos^{-1}\frac 2 x))^2}}{\cos(\cos^{-1}\frac 2 x))^2}$$ $$=\frac{\sqrt{1-\frac 4 {x^2}}}{\frac 2 x}$$ $$=\frac{\sqrt{1-\frac 4 {x^2}}}{\frac 2 x}=\frac{x\sqrt{1-\frac 4 {x^2}}} 2=\frac{\sqrt{x^2-4}} 2$$

Which when multiplied by the $2$ on the outside of your integral gives the answer.

• I am suppose to have $\sqrt{x^2 - 4} - 2sec^{-1}(\frac{x}{2} +c$ – toby yeats Jun 4 '12 at 1:34
• How do you do x * $\sqrt{1 - \frac{4}{x^2}}$ – Jwan622 Feb 2 at 17:57

Multiply both top and bottom by $\sqrt{x^2 - 4}$ to obtain $$\int\dfrac{x^2 - 4}{x\sqrt{x^2 - 4}}\,\mathrm{d}x=\int\dfrac{x}{\sqrt{x^2 - 4}} + 4\int-\dfrac{1}{x}\dfrac{1}{\sqrt{x^2\left(1 - (2/x)^2\right)}}\,\mathrm{d}x$$ In the fisrt integral, set $u=x^2-4$ and $x\,\mathrm{d}x = \dfrac{1}{2}\,\mathrm{d}u$ to get $$\int\dfrac{x}{\sqrt{x^2-4}}\,\mathrm{d}x = \sqrt{x^2 - 4}+C_0.$$ In the second, consider $x>2$: $$2\int-\dfrac{2}{x^2}\dfrac{\mathrm{d}x}{\sqrt{1 - (2/x)^2}}$$ Now, set $u=\dfrac{2}{x}$ and $\mathrm{d}u = -\dfrac{2}{x^2}\,\mathrm{d}x$. Then, \begin{aligned} 2\int-\dfrac{2}{x^2}\dfrac{\mathrm{d}x}{\sqrt{1 - (2/x)^2}} &=2\int\dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} \\ &=2\arcsin u + C_1 \end{aligned}

Adding them up: $$\int\dfrac{\sqrt{x^2 - 4}}{x}\,\mathrm{d}x = \sqrt{x^2 - 4} + 2\arcsin\left(\frac{2}{x}\right) + C$$

Note that, if $x<-2$, we would obtain the integral $$\int\dfrac{\sqrt{x^2 -4}}{x}\,\mathrm{d}x = \sqrt{x^2 - 4}-2\arcsin\left(\dfrac{2}{x}\right)+C,$$ but, since $x<-2$, $-2\arcsin\left(\dfrac{2}{x}\right)>0$, and the sign would become positive.

Therefore, we can write \begin{aligned} \int\dfrac{\sqrt{x^2 - 4}}{x}\,\mathrm{d}x &= \sqrt{x^2-4}+2\arcsin\left(\dfrac{2}{|x|}\right) + C \\&= \sqrt{x^2-4}+2\operatorname{arccsc}\left(\dfrac{|x|}{2}\right) + C \end{aligned}

Note that the arbitrary constants are not necessarily the same.

Another answer to this question probably isn't necessary. This post is really just an excuse to introduce a technique I wish someone had taught me in high-school.

Suppose $x>2$. Substitute $\sqrt{x^2-4}=(x-2)t$. This is the third substitution of Euler. Note that we have $t>1$. Solving for $x$ in the substitution relation, we have $x=\frac{2(t^2+1)}{t^2-1}\implies\mathrm{d}x=-\frac{8t}{(t^2-1)^2}\,\mathrm{d}t$. This transforms the integrand to a rational function of $t$:

$$\int\frac{\sqrt{x^2-4}}{x}\,\mathrm{d}x=-\int\frac{16t^2}{(t^2+1)(t^2-1)^2}\,\mathrm{d}t.$$

Expanding via partial fractions,

$$-\frac{16t^2}{(t^2+1)(t^2-1)^2}=\frac{4}{t^2+1}-\frac{2}{(t-1)^2}-\frac{2}{(t+1)^2}.$$

Hence, the integral has been reduced to three elementary integrals, and we've done it without the usual machinery of trigonometric substitutions.