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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.

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

up vote 1 down vote accepted

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 $$\sqrt{x^2 - 4} - 2 \, \text{arsec}(x/2)) + c = \sqrt{x^2 - 4} - 2 \, \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right) + c$$

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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.

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I am suppose to have $\sqrt{x^2 - 4} - 2sec^{-1}(\frac{x}{2} +c$ –  user138246 Jun 4 '12 at 1:34
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