Relation between inverse tangent and inverse secant 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}}]


Which shows that the two answers differ by $\pi/12$, but only for $x>3$.
 A: 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,
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{1}{3}\sqrt{x^2-9}.$$
A: 
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 $\frac{\pi}{12}$.
A: 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

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

A: 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}. $ 

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