# 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.

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## Answers and Replies

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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

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