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

$x = 3 \sec \theta \implies dx = 3 \sec\theta \tan\theta d\theta$

$$\begin{align} \int \frac{dx}{\sqrt{x^2 -9}} & = \frac{1}{3}\int \frac{3 \sec\theta \tan\theta d\theta}{\tan\theta} \\ \\ & = \int \sec\theta d\theta \\ \\ & = \ln | \sec\theta + \tan\theta| + c\end{align}$$

$x = 3\sec \theta \implies \sec\theta = \frac {x}{3}$

$\tan\theta = \frac{\sqrt{x^2 - 9}}{x}$

I have have confused x with 3 but I cannot get the proper answer which is

$$\ln | x + \sqrt{x^2 - 9}| + c$$

I always get $\dfrac{x}{3}$ or $\dfrac{\sqrt{x^2 - 9}}{x}$ or some variation of that, I can't eliminate them to get their answer.

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What you should end with is

$\sec\theta = \dfrac x3\quad$ and $\quad\tan \theta = \dfrac{\sqrt{ x^2 - 9}}{3}$.

Then you have $$\begin{align} \log \Big| \frac 13\left(x + \sqrt{x^2 - 9}\right)\Big| + C & = \log|x +\sqrt{x^2 - 9}| -\log 3 + C \\ \\ & = \log|x + \sqrt{x^2 - 9}| + C'\end{align}$$

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  • $\begingroup$ My book doesn't have the 1/3, I have spent about an hour on this proble and I just can't figure out how they got it. $\endgroup$ – Sailor Jim Jul 4 '13 at 15:38
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    $\begingroup$ The factor $\frac 13$ in the log becomes a constant: $$\log\left(\frac ab\right) = \log a - \log b$$ $\endgroup$ – Namaste Jul 4 '13 at 15:39
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    $\begingroup$ That is a dirty trick. $\endgroup$ – Sailor Jim Jul 4 '13 at 15:43
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    $\begingroup$ I agree, you were simply off by a constant of integration. Keep a look out, this "dirty trick" will come up again and again! ;-) $\endgroup$ – Namaste Jul 4 '13 at 15:45
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    $\begingroup$ @SailorJim Be prepared, it is a constant trick in these matters ;-). (+1, amWhy). $\endgroup$ – Julien Jul 4 '13 at 15:46
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$\theta=arc\sec\frac{x}{3}\implies \sec\theta=\frac{x}{3}$ and then $\tan\theta=\frac{\sqrt{x^2-9}}{3}$ so, integral would be $\ln\left(\frac{x}{3}+\frac{\sqrt{x^2-9}}{3}\right)+c=\ln(x+\sqrt{x^2-9})+k(=-\ln 3+c)$

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  • $\begingroup$ How does the inner ln term cancel out their denominators? $\endgroup$ – Sailor Jim Jul 4 '13 at 15:33
  • $\begingroup$ i made some mistake there. See the updated answer. $\endgroup$ – Aang Jul 4 '13 at 15:38
  • $\begingroup$ What is k? That part confused me. $\endgroup$ – Sailor Jim Jul 4 '13 at 15:38
  • $\begingroup$ For indefinite integral constants doesn't matter much , you just collect all the contants and call it another final constant. So first constant of integration $c$ with another constant $-\ln 3$ is just another constant $k$. $\endgroup$ – Aang Jul 4 '13 at 15:41
  • $\begingroup$ Oh okay I see now. $\endgroup$ – Sailor Jim Jul 4 '13 at 15:44

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