2.9 Calculus of Hyperbolic Functions

[STARTS AT 19:00]

INSTRUCTOR: OK, so the next one we've got, x squared minus 4, well, our u is x. OK? And we have an a of 2. All right, so this is going to be, once we take the integral, hyperbolic cosine of x/2. OK. 

All right, next, what's being squared here? Well, how I could write this, I could write this as e to the x squared. So our u is e to the x, which makes our du e to the x dx. 

OK. Well, we've sort of got this redundant because our e to the x is actually a u, so I could also write this as du over u equals dx. OK. So the way this is going to be written is 1/u square root of 1 minus u squared du, which will be a negative secant hyperbolic secant inverse, absolute value of u plus c. And box that one in. Which in turn leaves secant hyperbolic inverse absolute value of e to the x. Now, e to the x is always positive. So I don't need those absolute value bars. I'm going to put that in parentheses, though

[ENDS AT 21:03]