Find the integral of (e^2x-e^-2x)/(e^2x+e^-2x) ???

Question

anonymous · Answer

Don't know where to start with this problem... Here's a drawing of it if it's more helpful. |dw:1394694885579:dw|

fibonaccichick666 · Answer

Are you familiar with the exponential forms of the trig functions?

anonymous · Answer

e^2x+e^-2x=t

anonymous · Answer

differntiate it first

fibonaccichick666 · Answer

If so, this becomes really easy

kainui · Answer

Since this is just tanh(2x) you can say the derivative is simply 2sech(2x)!

anonymous · Answer

Wait what.. I thought the integral of tan was -ln (cosu) +C?

fibonaccichick666 · Answer

wait wait, first let's get to tanh @Kainui

anonymous · Answer

e^2x+e^-2x=t
on differntiating
(2*e^2x -2*e^-2x)*dx=dt
2*(e^2x -e^-2x)*dx=dt
(e^2x -e^-2x)*dx=dt/2

fibonaccichick666 · Answer

and  dixie, not quite the same, if you are not familiar with the exponential form of trig fns.  @dixiemitsy , have you seen this before? $sinx=\frac{e^x-e^-x}{2}$

anonymous · Answer

Nope haven't seen it.. and @g17 does that mean you wouldn't have to take the integral of the bottom because it would cancel out?

fibonaccichick666 · Answer

ok then the approach that everyone except g17 is taking, will not make sense to you

anonymous · Answer

Ok, thanks :P

fibonaccichick666 · Answer

Do you understand g17's approach?

anonymous · Answer

Yup, got it now :))

fibonaccichick666 · Answer

alrighty good luck! :)

fibonaccichick666 · Answer

Just to make me feel better, what did you get?