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I've been working on the following integral $$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$ where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large number of steps I achieved the correct answer: $$\int\frac{\sqrt{x^2-9}}{x^3}\,dx=\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}+C$$

I was able to check my answer using Mathematica.

expr = D[1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(2 x^2), x];
Assuming[x >= 3, FullSimplify[expr]]

Which returned the correct response:

Sqrt[-9 + x^2]/x^3

Mathematica returns the following answer:

Integrate[Sqrt[x^2 - 9]/x^3, x, Assumptions -> x >= 3]

-(Sqrt[-9 + x^2]/(2 x^2)) - 1/6 ArcTan[3/Sqrt[-9 + x^2]]

Which I can write to make more clear.

$$-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}+D$$

Now, you can see that part of my answer is there, but here is my question. How can I show that $$\frac16\sec^{-1}\frac{x}{3}\qquad\text{is equal to}\qquad -\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}$$ plus some arbitrary constant? What identities can I use? Also, can anyone share the best web page for inverse trig identities?

Update: I'd like to thank everyone for their help. The Trivial Solution's suggestion gave me: $$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$ Then the following identity came to mind: $$\tan^{-1}x+\tan^{-1}\frac1x=\frac{\pi}{2}$$ So I could write: \begin{align*} \frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2} &=\frac16\tan^{-1}\frac{\sqrt{x^2-9}}{3}-\frac{\sqrt{x^2-9}}{2x^2}\\ &=\frac16\left(\frac{\pi}{2}-\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)-\frac{\sqrt{x^2-9}}{2x^2}\\ &=\frac{\pi}{12}-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2} \end{align*} Using Olivier's and Miko's thoughts, I produced this plot in Mathematica.

Plot[{1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(
   2 x^2), -(1/6) ArcTan[3/Sqrt[x^2 - 9]] - Sqrt[x^2 - 9]/(
   2 x^2)}, {x, -6, 6},
 Ticks -> {Automatic, {-\[Pi]/12, \[Pi]/12}}]

enter image description here

Which shows that the two answers differ by $\pi/12$, but only for $x>3$.

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  • $\begingroup$ the arbitrary constant may be $\frac\pi{12}$ (judging from a plot, valid for $x\ge3$ only) $\endgroup$
    – Mirko
    Jan 4, 2017 at 7:07
  • $\begingroup$ Lookup Wikipedia. $\endgroup$
    – user65203
    Jan 4, 2017 at 8:08

4 Answers 4

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What you are asking to prove is incorrect, I believe.

By the substitution, we have that $$\frac{x}{3}=\sec(\theta)\Leftrightarrow\frac{3}{x}=\cos(\theta).$$

By the Pythagorean identity,

$$\sin(\theta)=\sqrt{1-\frac{9}{x^2}}=\sqrt{\frac{x^2-9}{x^2}}.$$

Therefore,

$$ \tan(\theta)=\sqrt{\frac{x^2-9}{x^2}}\frac{x}{3}=\frac{1}{3}\sqrt{x^2-9}$$

Hence, $$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{1}{3}\sqrt{x^2-9}.$$

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Hint. Applying the chain rule, for $x>3$, one may check that $$ \left(\frac16\sec^{-1}\frac{x}{3}\right)'=\frac{1}{2x\sqrt{9-x^2}} $$$$ \left(-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)'=\frac{1}{2x\sqrt{9-x^2}} $$ since $$ \lim_{x \to 3^+}\left(\frac16\sec^{-1}\frac{x}{3}\right)=0,\quad \lim_{x \to 3^+}\left(-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)=-\frac{\pi}{12} $$ then

$$ \frac16\sec^{-1}\frac{x}{3}=-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}+\frac{\pi}{12},\quad x>3. $$

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  • $\begingroup$ I think your second limit should be $-\pi/12$ and the final answer should have a plus $\pi/12$. $\endgroup$
    – David
    Jan 5, 2017 at 1:57
  • $\begingroup$ @David Typo edited. Thank you very much. $\endgroup$ Jan 5, 2017 at 2:09
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enter image description here

Pulled straight off wikipedia.

You seem to have used ** intersection of tan($\theta$) and arcsec(x) **

Also, that "constant" you were referring to is probably $\frac{\pi}{12}$.

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Since

$$ \sec[..]^2 = 1+ \tan[..]^2 $$

we directly have identities

$$ \sec^{-1}x = \tan^{-1}\sqrt {x^2 - 1} $$

and

$$ \tan ^{-1} x= \sec^{-1}\sqrt {1 + x^2} $$

Also plot between $ \sec^{-1}...,\, \tan^{-1}. $

enter image description here

Note the $ \pi/4,\pi $ intercept on $y$, and period respy.Proper sign to be taken.

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  • $\begingroup$ sorry for typos.. $\endgroup$
    – Narasimham
    Jan 4, 2017 at 8:28

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