Integration using the hyperbolic functions

**chella182** · May 5th 2009, 11:39 AM

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

the question goes:

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

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

**HallsofIvy** · May 5th 2009, 12:10 PM

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

**chella182** · May 5th 2009, 12:12 PM

LaTeX fail

**HallsofIvy** · May 5th 2009, 12:16 PM

Edited.

**chella182** · May 5th 2009, 12:20 PM

Thankies.

**chella182** · May 6th 2009, 09:03 AM

Related to my calculations with the original question.

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

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