How does $\sec^{-1}\left(\frac{x}{3}\right)$ turn out to be $\frac 13\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)$?

Question

Solve using trig substitution:

$\int\left(\frac{dx}{x\sqrt{x^{2}-9}}\right)$

1. $u = x = \left({3\sec\theta}\right)^{2} = 9\sec^{2}\theta$ and $ dx = \sec\theta\tan\theta$
2. $\theta = \sec^{-1}\left(\frac{x}{3}\right)$ and $ \sqrt{x^2-9} = 3\tan\theta$
3. $\int \frac{\sec\theta\tan\theta}{9\sec\theta\tan\theta}d\theta$
4. $\frac{1}{9}\int d\theta = \frac{1}{9}\theta = \frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C$

As you can see, we end up at $$ \bbox[5px,border:2px solid #282] {\frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C} $$

However, when I use calculators to solve the same problem, I end up getting $$ \bbox[5px,border:2px solid red] {\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) + C} $$ or $$ \bbox[5px,border:2px solid red] {\sqrt{x^2-9} + 3tan^{-1}\left(\frac{3}{\sqrt{x^2-9}}\right) + C} $$

WolframAlpha does not explain why (obviously I don't know) $ \sec^{-1}x$ ended up being simplified to $\tan^{-1}x$, the $ x $ parameter being whatever with substitute in for.

My guess is it has to do with the trig substitution rules.

I haven't found any trig identity that looks like that or why the answer is simplified this way, and differently if that matters at all.

Can someone explain why?Thanks.

Answer 1

Try making a triangle. with hypotenuse $x$, base $3$. You will get the height to be $\sqrt{x^2-9}$. So, wrt to that angle, $$\sec^{-1}(\frac{x}{3})=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$

Answer 2

From

$$\cos^2t+\sin^2t=1$$ you draw

$$1+\tan^2t=\sec^2t,$$

$$\tan t=\sqrt{\sec^2t-1}.$$ Then with $t=\text{arcsec }x$, taking the arc tangent,

$$t=\arctan\sqrt{x^2-1}=\text{arcsec }x.$$

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Similarly,

$$\text{arcsec}\sqrt{x^2+1}=\arctan x.$$

Answer 3

Let us prove that : $$\sec^{-1}\left(\frac{x}{3}\right)=\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)$$

Starting from L.H.S., put $x=3 \sec t$, you'll get :

$$L.H.S.=\sec^{-1}\left(\frac{x}{3}\right)=t$$

Now do the same with R.H.S.

$$R.H.S.= \tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)=\tan^{-1}\left( \sqrt{\sec^2t-1 }\right)=\tan^{-1}({\tan t})=t$$

Therefore $$L.H.S=R.H.S$$

Answer 4

$\sec^{-1}(x/3) = \alpha \\ \therefore \sec(\alpha) = 1/\cos(\alpha) = x/3 => \cos(\alpha) = \dfrac{3}{x}$

$\tan(\alpha)=\dfrac{\sqrt{1-\cos^2(\alpha)}}{\cos(\alpha)}=\dfrac{\sqrt{1-\dfrac{9}{x^2}}}{\dfrac{3}{x}}=\dfrac{\sqrt{x^2-9}}{3}$

$\therefore \tan^{-1}(\tan(\alpha))=\alpha=\sec^{-1}(x/3)=\tan^{-1}(\dfrac{\sqrt{x^2-9}}{3})$