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Attempting integral:

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

Let $x = 3\ sec\ \theta$ so that under the square root we have:

$$\sqrt{9\ sec^2\ \theta - 9}$$

$$\sqrt{9(sec^2\ \theta - 1)}$$

$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$

The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:

$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$

The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:

$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$

Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:

$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$

However, using an online calculator, the answer turned out to be:

$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$

It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that

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

becomes $$\sqrt{u^2 - 1}$$

And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?

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Your mistake:

$$\sec\theta=\frac x3$$ yields

$$\cos\theta=\frac 3x,\\\sin\theta=\sqrt{1-\frac9{x^2}},\\\tan\theta=\sqrt{\frac{x^2}9-1}.$$

Not $\sqrt{x^2-9}$.


This said, you can rescale the variable with $x=3t$ to get

$$\int\frac{dt}{\sqrt{t^2-1}}.$$

You may recognize another familiar integral,

$$\int\frac{dt}{\sqrt{1-t^2}}=-\arccos t+C.$$

What you have here is just the hyperbolic equivalent,

$$\int\frac{dt}{\sqrt{t^2-1}}=\text{arcosh }t+C,$$

which can also be written

$$\log\left(t+\sqrt{t^2-1}\right)+C.$$

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HINT:

If $x=3\sec\theta$

$$\tan^2\theta=\left(\dfrac x3\right)^2-1=\dfrac{x^2-9}9$$

$|\tan\theta|=\dfrac{\sqrt{x^2-9}}3$

$\sec\theta+|\tan\theta|=\dfrac{x+\sqrt{x^2-9}}3$

$\implies\ln(\sec\theta+|\tan\theta|)=\ln(x+\sqrt{x^2-9})-\ln3$

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  • $\begingroup$ Oh right, so I forgot that $1/3$ is part of the expression for $tan\ \theta$ here, and that'll cancel out the $1/3$ in both terms in the solution? And the $ln\ 3$ bit wasn't in the solution on the calculator.. odd $\endgroup$ – sangstar Jul 11 '17 at 7:05
  • $\begingroup$ @sangstar, As $$-\ln3+C=C'$$ But don't miss the modulus. $\endgroup$ – lab bhattacharjee Jul 11 '17 at 7:06
  • $\begingroup$ Your answer makes sense and I appreciate the help, however I have two quick questions. a) Modulus because $\tan\ \theta$ must be larger than $\sec\ \theta$ for all $\theta$? b) I find it odd that integrating an indefinite integral spits out the constant $ln\ 3$. In this case, what influence is $ln\ 3$ doing to $C'$ to make it differ from $C$? $\endgroup$ – sangstar Jul 11 '17 at 7:52
  • $\begingroup$ @sangstar, $\tan\theta$ can be $<0$ unlike $$\dfrac{\sqrt{x^2-9}}3$$ right? Difference of constant is constant right? $\endgroup$ – lab bhattacharjee Jul 11 '17 at 8:04
  • $\begingroup$ That's right. And yeah, of course, I've just never had a constant come out from a definite integration before I don't think. $\endgroup$ – sangstar Jul 11 '17 at 8:06
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$$\int\frac{dx}{\sqrt {x^2-9}}$$

By substitution

$$x=3\sec\theta $$

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

$$\int \frac{3\sec\theta \tan \theta d\theta}{\sqrt{9\sec^2\theta-9}} $$

$$\int \frac{3\sec\theta \tan \theta d\theta}{3\tan\theta} $$

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

$$ ln|\sec\theta+\tan\theta|$$

$$ ln| \frac{1+\sin \theta}{cos \theta}|$$

$$ \cos \theta=\frac{3}{x}$$

Recall that $\sin \theta= \sqrt{1-\cos^2\theta}$

$$ln|\frac{x(1+\sqrt{1-\frac{9}{x^2})}}{3}|$$

$$ ln|x+\sqrt{x^2-9}|-ln3$$

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$$I = -\int \frac{1}{\sqrt{x^2-9}}\,dx$$ $\sqrt{bx^2 - a} \implies x = \sqrt{\frac{a}{b}}\sec \alpha$ $$x = 3\sec\alpha \implies \frac{dx}{d\alpha}=3\sec\alpha\tan\alpha$$

$$ -\int \frac{1}{\sqrt{\left(3\sec \alpha \right)^2-9}}3\sec\alpha\tan\alpha\,\,d\alpha$$ $$ -\int \frac{\sec\alpha \tan\alpha}{\sqrt{\sec^2\alpha-1}}\,d\alpha$$ $\sec^2\alpha = 1 + \tan^2\alpha$ $$ -\int \frac{\sec\alpha \tan\alpha}{\sqrt{\tan^2\alpha}}\,d\alpha = -\int \frac{\sec\alpha \tan\alpha}{\tan\alpha}\,d\alpha = -\int\sec\alpha\,d\alpha$$ $$-\int\sec\alpha\,d\alpha = -\ln(\tan\alpha + \sec\alpha) +C$$ $$\alpha = arcsec\left( \frac{1}{3}x\right)$$ $$-\ln\left(\tan\left( arcsec\left( \frac{1}{3}x\right)\right)+ \sec\left( arcsec\left( \frac{1}{3}x\right)\right)\right) +C$$

$$-\ln\left(\sqrt{\frac{1}{9}x^2 -1} + \frac{1}{3}x\right)+C$$

$$-\ln\left(\frac{1}{3} \left(\sqrt{x^2 - 9} + x\right)\right)+C$$

$$-\ln\left(\sqrt{x^2 - 9} + x\right) - \ln(3)+C$$

$$I = -\ln\left(\sqrt{x^2 - 9} + x\right) +C$$

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