## Evaluate the line integral

1. The problem statement, all variables and given/known data

Evaluate ∫(x^3 + y^3)ds where C : r(t)=<e^t , e^(-t)>, 0 <= t <= ln2
c

2. Relevant equations

3. The attempt at a solution

I tried to parametrize the integral and change ds to sqrt(e^(2t) + e^(-2t)) dt.

I then change (x^3 + y^3) to (e^(3t) + e^(-3t)

so i ended up with

ln2
∫(e^(3t) + e^(-3t)) * sqrt(e^(2t) + e^(-2t))dt
0

I feel like i set the integral up wrong becuase I would have no idea of how to do this integral. Even wolframalpha gives me a crazy answer. Is there another way to do this or did i make a mistake?
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 Recognitions: Gold Member Homework Help It looks correct to me. For what it's worth, Maple gives:$$\frac 1 8\ln \left( \frac{13\sqrt{17}+51}{13\sqrt{17}-51}\right)+\frac{63\sqrt{17}}{64}$$
 So i have to actually do that integral? Is there a way to write it in differential form or another form that would be easier?

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Homework Help

## Evaluate the line integral

 Quote by meson0731 So i have to actually do that integral? Is there a way to write it in differential form or another form that would be easier?
Beats me. I don't right off see a simple way to work it myself. I thought about expressing the integrand in terms of ##\cosh(3t)## and ##\cosh(2t)## and I still didn't see anything obvious. But then again, I haven't been losing any sleep over it and maybe someone else will see something clever.
 Yeah i tried changing it into hyperbolic but it just got even more messy....
 It may help to rewrite the curve that it travels over. Instead of integrating $$r(t)=$$ it may be easier to integrate $$r(t)=$$ It haven't tried it though, so it may not be any easier.
 I'm working on this same problem. I did indeed rewrite it as r(t)=, but this integral is no easier to solve. The solution given by Wolfram Alpha for this integral was the same numerically as that given by Maple in the above post.
 I haven't evaluated many line integrals, but if we are on the curve y = 1/x, can't we skip using the parameter t and integrate $(x^3 + \frac{1}{x^3})\sqrt{1 + \frac{1}{x^4}} dx$ from 1 to 2? I'm having success evaluating it using the substitution $x^2 = tan\theta$.