# Hyperbolic functions

• Mar 19th 2010, 08:31 AM
jpquinn91
Hyperbolic functions
I have to find the integral of 1/((25x^2-49)^1/2) using hyperbolic functions. I'm not sure how to start to know which functions to use, thanks
• Mar 19th 2010, 08:46 AM
drumist
It sounds like you are expected to know (or be able to determine from a table) the derivatives of the inverse hyperbolic functions. In particular, the one that is useful in this problem would be:

$\displaystyle \frac{d}{dx} \cosh^{-1}(x) = \frac{1}{\sqrt{x^2-1}}$

(for $\displaystyle x>1$)

Can you solve this with this information?
• Mar 19th 2010, 10:35 AM
jpquinn91
I wrote x as 7cosh(u) and ended up with arccosh(x/2) between 7 and 14, i'm not sure that this is right?
• Mar 19th 2010, 11:29 AM
drumist
You didn't say what the original endpoints were.

I'm not sure which simplification method you use, but this is how I get it:

$\displaystyle \int \frac{1}{\sqrt{25x^2-49}} dx$

$\displaystyle =\frac{1}{7} \int \frac{1}{\sqrt{\tfrac{25}{49}x^2-1}} dx$

$\displaystyle =\frac{1}{7} \int \frac{1}{\sqrt{\left(\tfrac{5}{7}x\right)^2-1}} dx$

Let $\displaystyle u=\tfrac{5}{7}x$ and $\displaystyle du=\tfrac{5}{7}dx$.

$\displaystyle =\frac{1}{5} \int \frac{1}{\sqrt{u^2-1}} du$

$\displaystyle =\tfrac{1}{5} \cosh^{-1} u + C$

$\displaystyle =\tfrac{1}{5} \cosh^{-1}\left(\tfrac{5}{7}x\right) + C$

And this is valid for $\displaystyle u>1 \implies x>\tfrac{7}{5}$.