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$$\int \frac{dx}{\sqrt{x^2 - 9}}$$

$x = 3\sec\theta$

$dx = 3\tan\theta \sec\theta\,d\theta$

$$\frac{1}{3} \int \frac{3\tan\theta\sec\theta}{\sqrt{\sec^2 + 1}} d\theta$$

$$ \int \sec\theta d\theta$$

I have no idea how to continue without looking up tables which I cannot do on a test, how would I proceed or have I already gone wrong?

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Try a similar trick to the one used in here. – Git Gud Jul 2 '13 at 23:34
I am not clever enough to make that work. What is the goal? – Dantheman Jul 2 '13 at 23:37
Stop the self deprecating comments. Mathematics beats down EVERYONE. The goal is the same as in the link I posted above. Did you understand what happened in the other question? – Git Gud Jul 2 '13 at 23:39
Not entirely, I don't see the goal of putting in what seems to be random trig fucntions. – Dantheman Jul 2 '13 at 23:41
I added more stuff to my answer. – Git Gud Jul 2 '13 at 23:47

Hint: Multiply $\sec$ by $\displaystyle 1=\frac{\sec(\theta)+\tan (\theta)}{\sec (\theta)+\tan (\theta)}$.

Further hint: $\displaystyle \int \sec (\theta) d\theta=\int \sec (\theta) \cdot 1\,d\theta=\int \sec (\theta) \frac{\sec(\theta)+\tan (\theta)}{\sec (\theta)+\tan (\theta)}d\theta=\int \frac{(\sec(\theta))^2+\sec (\theta)\tan (\theta)}{\sec (\theta)+\tan (\theta)}d\theta$.

Now differentiating $\theta \mapsto \sec (\theta) + \tan (\theta)$, you can find $\displaystyle \frac{d}{d\theta}\left(\sec (\theta) + \tan (\theta)\right)=(\sec(\theta))^2+\sec (\theta)\tan (\theta)$.

Finally use $\displaystyle \int \frac{u'}{u}=\log (|u|).$

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That doesn't seem right, how do I differentiate that? – Dantheman Jul 2 '13 at 23:52
@Dantheman I added more stuff to my answer. I gotta run. Hopefully someone else will assist you if you need it. Otherwise I'll check back in a few hours. – Git Gud Jul 2 '13 at 23:57

Another method: \begin{eqnarray*} \int\sec\theta\mathrm{d}\theta&=&\int\frac{\cos\theta}{\cos^2\theta}\mathrm{d}\theta=\int\frac{\mathrm{d}\sin\theta}{(1-\sin\theta)(1+\sin\theta)}\\ &=&\frac{1}{2}\left(\int\frac{\mathrm{d}\sin\theta}{1-\sin\theta}+\int\frac{\mathrm{d}\sin\theta}{1+\sin\theta}\right)=\cdots \end{eqnarray*}

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This method is the most straightforward one posted so far. – Michael Hardy Jul 3 '13 at 0:28
Where does the first sin come from? – Dantheman Jul 3 '13 at 0:31
@Dantheman $\mathrm{d}\sin\theta=\cos\theta\mathrm{d}\theta$ – Cezhong-Tong Jul 3 '13 at 0:40
What is d? I domt follow – Dantheman Jul 3 '13 at 0:50
@Dantheman Consider $\sin (\theta)$. Differentiate it: $\displaystyle \frac{d\sin (\theta)}{d\theta}=\cos (\theta)$. Now the wrong part comes, (but it works). 'Multiply' by $d\theta$ to get $d\sin (\theta)=\cos (\theta)d\theta$. It does not only work in this case, it works with integrals in general. It is formally wrong, but it works. – Git Gud Jul 3 '13 at 8:28

I will show you the work with Euler's substitutions so you see that you don't need to come up with any imaginative trick.

First the Euler's substitutions tells you to introduce a new variable $t$ such that $\sqrt{x^2-9}=x+t$. From this we get that $x^2-9=x^2+2xt+t^2$, i.e. $$x=\frac{-9-t^2}{2t}$$ and $$\sqrt{x^2-9}=\frac{-9-t^2}{2t}+t$$

Then, differentiating you get $$dx=\left(\frac{9}{2t^2}-\frac{1}{2}\right)\text{d}t$$.

Putting these in the integral we get

$$\int \frac{\text{d}x}{\sqrt{x^2-9}}=\int\frac{2t}{-9+t^2}\frac{9-t^2}{2t^2}\text{d}t.$$

Now simplify a little to get $$\int\frac{-\text{d}t}{t}=-\ln(t).$$

Returning now to the old variable $x$, using that $t=\sqrt{x^2-9}-x$ we get


If you want you can also rewrite this in the equivalent form: $$\ln\left(\frac{1}{\sqrt{x^2-9}-x}\right)=\ln\left(\frac{\sqrt{x^2-9}+x}{-9}\right)=\ln\left(\sqrt{x^2-9}+x\right)+constant.$$

I am telling you, those trigonometric substitutions are an educational nonsense. The only thing we had to do here was to notice that this was an integral of the form $R(\sqrt{ax^2+bx+c},x)$, for $R$ rational, and to remember what is the Euler substitution corresponding to it (there are only $3$ of them). The rest is just mindless computation.

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OK, just for fun, here's a substitution that works: $$ w = \tan\left(\frac\theta2+\frac\pi4\right) $$ If you're good at trigonometric identities, you can then derive these: $$ \sec\theta = \frac{1+w^2}{2w}, \qquad d\theta=\frac{2\,dw}{1+w^2}. $$ Then you have $$ \int\sec\theta\,d\theta = \int \frac{dw}{w} = \cdots\cdots $$ (I have a short note in the current Monthly about this. Also see the Wikipedia article about this integral, which I created a couple of years ago. You'll see a link to a paper about its application to cartography, that being the reason why people first wanted to know how to do this integral. This particular substitution is not (yet?) in the Wikipedia article.)

(What I privately think of as the cartographer's tangent half-angle formula says $\tan\left(\dfrac\theta2+\dfrac\pi4\right)=\sec\theta+\tan\theta$. But maybe someone else has called it that?)

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