Integrate?

Warrenx

(t^2 + e^(2t) + e^(-2t))^1/2

anyone got any ideas, I suck at integration so its hard for me to see what to do, if someone can start me in the right track I am sure I can finish.

thanks,

Warren.

p0oint

Hello!

Is the original integral that you want to solve:

$$\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt$$

Warrenx

Hello!

Is the original integral that you want to solve:

$$\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt$$

Yes that is it. I am sorry I did not use Latex, have headache right now and couldn't think

Warrenx

Dear Warrenx,

This integration could not be evaluated using elementary functions. You can get the answer using, Function calculator
hmm... I think I wrote the problem write:

It is asking for the arc length:

so s(t) =$$\displaystyle \int_0^1 |r'(t)|$$

r(t) = $$\displaystyle \sqrt2t , e^t , e^{-t}$$

so I get r'(t) = $$\displaystyle t , e^t , e^{-t}$$

then lr'(t)l = $$\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}$$

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck

simplependulum

MHF Hall of Honor
hmm... I think I wrote the problem write:

It is asking for the arc length:

so s(t) =$$\displaystyle \int_0^1 |r'(t)|$$

r(t) = $$\displaystyle \sqrt2t , e^t , e^{-t}$$

so I get r'(t) = $$\displaystyle t , e^t , e^{-t}$$

then lr'(t)l = $$\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}$$

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck

Oh , the derivative of $$\displaystyle \sqrt{2} t$$ is $$\displaystyle \sqrt{2}$$

so what the integral you are going to evaluate should be :

$$\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt$$

Trick is also found

$$\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2$$ (Happy)

the heaven is just in front of you now .

Warrenx

Warrenx

Oh , the derivative of $$\displaystyle \sqrt{2} t$$ is $$\displaystyle \sqrt{2}$$

so what the integral you are going to evaluate should be :

$$\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt$$

Trick is also found

$$\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2$$ (Happy)

the heaven is just in front of you now .
If you don't tell anyone about this, I wont tell anyone...
but thank you!