# Very short math question integral?

So our teacher solved this integral and in the end it was something like $\sqrt{x^2-4} - 2\operatorname{arcsec}\left(\frac{x}{2}\right)$ and my teacher said this is equivalent to $\sqrt{x^2-9} + 2\arctan\left(\frac{2}{\sqrt{x^2-4}}\right)$ ..why? I dont understand how $\operatorname{arcsec}\left(\frac{x}{2}\right)$ can be equal to $\arctan\left(\frac{2}{\sqrt{x^2-4}}\right)$?

• Does arctg mean $\arctan$? – ferson2020 Jan 30 '13 at 19:24
• yes it does mean arctan – user60290 Jan 30 '13 at 19:25
• Draw a right triangle, with angle $\sec^{-1}(x/2)$, by labelling the hypotenuse $x$, the adjacent side $2$. The remaining side is $\sqrt{x^2-4}$. (There is a potential sign issue when angle is not between $0$ and $\pi/2$, so I am being a bit sloppy.) – André Nicolas Jan 30 '13 at 19:26
• (1) Please do use LaTeX to write mathematics in this site. The FAQ section gives some directions ; (2) What integral? – DonAntonio Jan 30 '13 at 19:28
• sqrt(x^2-4)/x.. – user60290 Jan 30 '13 at 19:29

$\operatorname{arcsec} \frac x2$ is the angle whose secant is $\frac x2$, which is the same as $\arccos \frac 2x$. If you draw a right triangle with the side adjacent to angle $A$ $2$ and hypotenuse $x$, the opposite side is then $\sqrt {x^2-4}$, then $\tan A=\frac {\sqrt {x^2-x^4}}2$ and $\cot A=\frac 2{\sqrt {x^2-x^4}}$. Are you sure it wasn't $\operatorname{arcsec} \frac x2=\operatorname{arccot} \frac 2{\sqrt {x^2-x^4}}$? There are many such relations between the inverse trig functions. I don't remember them, I draw the triangle as described and find them as needed.