# Thread: Integration using the hyperbolic functions

1. ## Integration using the hyperbolic functions

Okay, I know this should be easy, but I'm having right bother with it the question goes:

Find $\displaystyle \int\frac{dx}{\sqrt{4x^2-9}}$

All I know is I need to find the right substitution. Can anyone help?

2. Since you mention hyperbolic functions, $\displaystyle cosh x= \frac{e^x+ e^{-x}}{2}$, and $\displaystyle sinh x= \frac{e^x- e^{-x}}{2}$ so $\displaystyle cosh^2 x= \frac{e^{2x}+ 2+ e^{-2x}}{4}$ while $\displaystyle sinh^2 x= \frac{e^{2x}- 2- e^{-2x}}{4}$. If we subtract the second from the first, all terms involving $\displaystyle e^{2x}$ cancel leaving $\displaystyle cosh^2 x- sinh^2 x= 1$ or $\displaystyle cosh^2 x- 1= sinh^2 x$. In order to use that with $\displaystyle \sqrt{4x^2- 9}$ you want to make those look alike. One way to do that is factor out 9: $\displaystyle \sqrt{9((4/9)x^2-1}= 3\sqrt{(4/9)x^2- 1}$. Now the substitution cosh t= (2/3)x gets rid of the square root.

3. LaTeX fail

4. Edited.

5. Thankies.

6. ## Question about taking out a constant from an integral

Related to my calculations with the original question.

If I have...
$\displaystyle \frac{1}{3}\int\frac{\frac{3}{2}\sinh{u}}{\sqrt{(\ cosh{u})^{2}-1}}du$
...when I take the $\displaystyle \frac{3}{2}$ out, will it become $\displaystyle \frac{1}{2}$ in front by $\displaystyle \frac{1}{3}\times\frac{3}{2}$, or will it become $\displaystyle \frac{9}{2}$ by $\displaystyle \frac{\frac{3}{2}}{3}=\frac{9}{2}$?