Recent content by abouchard

Okay, scratch that. After finding substitutions for both \sinh ^2 and \cosh ^2, I got this: \frac{9}{2} \int \cosh (2x) + 1 dx If I remember my rules correctly (and I'm not doing anything stupid), this is \frac{9}{4} \sinh (2x) + x Is there a better way to simplify this than...

I hate to dig this back up, but I'm trying to do something similar to this, and I'm not having much luck. I don't know if it's because I'm not that familiar with hyperbolic trig functions or just rusty on integration. Using the trig substitution suggested with the equation provided, I get...