Hyperbolic functions

**jpquinn91** · March 19th 2010, 08:31 AM

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

**drumist** · March 19th 2010, 08:46 AM

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:

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

(for $x>1$ )

Can you solve this with this information?

**jpquinn91** · March 19th 2010, 10:35 AM

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?

**drumist** · March 19th 2010, 11:29 AM

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:

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

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

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

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

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

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

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

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

Thread: Hyperbolic functions

LinkBack

Thread Tools

Display