# Integration of e

## Homework Statement

I have the problem and answer, I'm just confused on the last step, but I'll put down everything anyways.

Find the arclength of the curve r(t) = <5√(2)t, e5t, e-5t>
0≤t≤1

L = ∫|r'(t)|

## The Attempt at a Solution

r'(t) = √((5√(2))2+(5e5t)2+(-5e-5t)2)

|r'(t)| = 5√(2+e10t+e-10t)

Now this is where I get confused.

∫(5√(2+e10t+e-10t)) = ((e10t-1)*√(2+e10t+e-10t))/(e10t+1)

I just don't understand how to integrate my answer to get that. If someone could just go through the steps that'd be great. Thanks!

Then obviously sub in for 1 and 0, getting 148.4064 which is correct so I know that the integration is correctly done.

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
That's an odd form for the resulting integral. To see how to do it more easily what is (e^(5t)+e^(-5t))^2? Expand it out.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

I have the problem and answer, I'm just confused on the last step, but I'll put down everything anyways.

Find the arclength of the curve r(t) = <5√(2)t, e5t, e-5t>
0≤t≤1

L = ∫|r'(t)|

## The Attempt at a Solution

r'(t) = √((5√(2))2+(5e5t)2+(-5e-5t)2)

|r'(t)| = 5√(2+e10t+e-10t)

Now this is where I get confused.

∫(5√(2+e10t+e-10t)) = ((e10t-1)*√(2+e10t+e-10t))/(e10t+1)

I just don't understand how to integrate my answer to get that. If someone could just go through the steps that'd be great. Thanks!

Then obviously sub in for 1 and 0, getting 148.4064 which is correct so I know that the integration is correctly done.
Your function $5\sqrt{2 + e^{10t} + e^{-10t}}$ can be re-written, using the identity $\cosh(10t) =2 \cosh(5t)^2 - 1,$ to give a much simpler integral.

RGV

Last edited: