1. ## Integration using identities

Using the identity, sinhx = 1/2(e^x-e^-x) and the substitution v=e^-x to find the integral of cosechx.

2. So you can't use the definition of cosh(x)?

3. You probably can, but I'm unsure of how to integrate a fraction.

4. Hello,

$\text{cosech}(x)=\frac{1}{\sinh(x)}$

that should finish the preliminaries

you'll have $2 \int \frac{1}{e^x-e^{-x}} ~dx$
substitute, as you're told, $v=e^{-x}$ (note that $e^x=\frac 1v$) and show your work...

5. Originally Posted by Erghhh
You probably can, but I'm unsure of how to integrate a fraction.
Is that a yes you can use it or yes you cannot?

edit: Moo - you showed me up! I forgot that simple identity. Good job.

6. Yes you can.

7. Is there a formula/general rule for integrating fractions?

8. Originally Posted by Erghhh
Is there a formula/general rule for integrating fractions?
You're making this too hard. "Fractions" is not a good way to think of this. For instance integrating $\frac{1}{x^2+2x+5}$ is integrating a fraction but is very complex. In your problem you have variables over a constant. Notice that $\frac{abc....z}{2}=\frac{1}{2} abc...z$. And you should know that any constant in front of an integral can be brought outside of the integral and applied to the answer after integration. So for this one, your constant is 1/2, bring it out and apply it after.

9. But am i not integrating 1/(v-v^-1)?

10. Originally Posted by Erghhh
But am i not integrating 1/(v-v^-1)?
Yes, sorry. But my thought still is true. There is general rule. I'll post with how to do it after I write this.

11. So like Moo said, v = e^{-x}, thus 1/v = e^{x}. Now use v-substitution. So we have this expression, $\frac{2}{\frac{1}{v}-v}$ If v=e^{-x}, $\ln(v)=-x$. Thus $\frac{1}{v}dv=-dx$. The new integral is now $-\int \frac{2}{\frac{1}{v}-v}\frac{1}{v}dv=\frac{2}{1-v^2}dv$. Are you with me so far?

12. And from there, you just use the standard form? so it's ln|1+v/1-v|?