Indefinite Integral Question; e^x

[Celadon]

Homework Statement

Find the indefinite integral (preferably using u-substitution):
∫(e^x-e^-x)² dx

Homework Equations

N/A

The Attempt at a Solution

To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be using u to substitute here. This is one of my attempts at solving the question:

∫[(e^x-e^-x)² dx]
= ∫[e^2x dx] - ∫[e^-2x dx]
= (e^2x / 2) + (1 / 2e^2x) + C

Could someone please tell me if I have the right answer / I'm on the right track / I'm completely wrong? If the latter, I'd appreciate it if you could help get me on track!

Thanks in advance!

 

[BruceW]

You almost got it. But you've forgotten the cross-terms when you did the square.

There is a good substitution you could use instead. But your method is just as good.

Edit: p.s. welcome to physicsforums, your name is good. reminds me of the town in pokemon

 

[rock.freak667]

If you need the u-substitution, try just putting u=e^x.

(Remember that 1/a is the same as a^-1)


EDIT: Also in your attempt you should also have


∫e^2x dx + ∫e^-2x dx +∫2 dx

 

[Celadon]

Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha.

Now then...
@BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh.
@rock.freak667: Ugh, how could I forget 1/a = a^-1? Thanks for the reminder. I have a question about the +∫2 dx, though. How exactly did it get there?

I think I'm actually a bit more confused than I initially thought I was. Let's say I went with this method:

u=e^x
du=e^xdx ; dx=du/e^x

...and then what? This is what I'm leaning towards:
=∫(u² dx) - ∫(1/u² dx)

Do I substitute dx for du/e^x or am I over-complicating things? Oh man, my head's going to burst. @@

 

[SammyS]

(e^x-e^-x)² = (e^x-e^-x)(e^x-e^-x)

Multiply it out. (Use FOIL, if that's what you're used to.)

 

[Dick]

(e^x-e^(-x))^2=(e^x-e^(-x))*(e^x-e^(-x)). Multiply that out and you'll see where the '2' came from. rock.freak667 had the sign wrong on it.

 

[Celadon]

Okay, do I tried expanding it and then went through this:

(e^x-e^-x)2 = (e^x-e^-x)(e^x-e^-x) = e^2x + e^-2x

∫(e^2x + e^-2x) dx
u = e^2x
du = (e^2x / 2) dx

Final solution:
(u²/e^2x) + (2 / e^2x)ln|e^2x|

 

[Mark44]

Okay, do I tried expanding it and then went through this:

(e^x-e^-x)2 = (e^x-e^-x)(e^x-e^-x) = e^2x + e^-2x

No, this is incorrect, as Sammy and Dick pointed out. (a - b)² ≠ a² - b², which is essentially what you are doing.
When you expand (e^x - e^-x)² you should get three terms, not two.

Also, there is an error below - you shouldn't get any log terms.

∫(e^2x + e^-2x) dx
u = e^2x
du = (e^2x / 2) dx

Final solution:
(u²/e^2x) + (2 / e^2x)ln|e^2x|

 

[Celadon]

Oh I see it now.
(e^x - e^-x)(e^x - e^-x) = e^2x - 2 + e^-2x

I somehow wrote "+ e⁰ - e⁰" at some point, oops. Thanks!
Time to try again. (:

 

[Mark44]

For you last error, I think this is what you did:

∫dx/e^2x = ln|e^2x|

That is incorrect. Don't write the integrand as a fraction - use the negative exponents.

 

[Celadon]

Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(e^x-e^-x)²dx = ∫(e^2x-2+e^-2x)dx
= ∫e^2xdx + ∫e^-2xdx - ∫2dx

Let u = e^x
du = e^xdx

∫e^2xdx + ∫e^-2xdx - ∫2dx = (1/e^x)(u³/3)+(1/e^x)(u^-1/-1)-2x+C
= (1/e^x)[(u³/3)+(u^-1/-1)]-2x+C
= (1/e^x)[(e^3x/3) - (1/e^x)]-2x+C
= (1/e^x)[(e^4x-3)/(3e^x)]-2x+C
= [(e^4x-3)/(3e^2x)]-2x+C

 

[SammyS]

Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(e^x-e^-x)²dx = ∫(e^2x-2+e^-2x)dx
= ∫e^2xdx + ∫e^-2xdx - ∫2dx

Let u = e^x
du = e^xdx

∫e^2xdx + ∫e^-2xdx - ∫2dx = (1/e^x)(u³/3)+(1/e^x)(u^-1/-1)-2x+C
= (1/e^x)[(u³/3)+(u^-1/-1)]-2x+C
= (1/e^x)[(e^3x/3) - (1/e^x)]-2x+C
= (1/e^x)[(e^4x-3)/(3e^x)]-2x+C
= [(e^4x-3)/(3e^2x)]-2x+C

What is [itex]\displaystyle\int du \,?[/itex]

 

[Celadon]

What is [itex]\displaystyle\int du \,?[/itex]

e^x+C?

 

[SammyS]

e^x+C?

Ignore my last post.

If u = e^x and du = e^xdx,

then [itex]\displaystyle\int e^{2x}dx[/itex] becomes what integral in terms of u ?

 

[Celadon]

Ignore my last post.

If u = e^x and du = e^xdx,

then [itex]\displaystyle\int e^{2x}dx[/itex] becomes what integral in terms of u ?

∫u²dx? :x

[EDIT] Another attempt:
2e^2x-2x-2e^-2x+C

x_x

 

[SammyS]

∫u²dx? :x

I think I'm confused as to what happens with du. ._.

If du = e^xdx, then dx = du(1/e^x), so:
(1/e^x)∫u²du?

Yes. You appear to be confused.

You can't take 1/e^x out of the integral. In fact you should leave it inside the integral as 1/u .

What is u²/u ?

 

[Celadon]

Yes. You appear to be confused.

You can't take 1/e^x out of the integral. In fact you should leave it inside the integral as 1/u .

What is u²/u ?

u¹ or just u? xD

(also, I edited my previous post with another answer in case you missed it.)

 

[SammyS]

u¹ or just u? xD

(also, I edited my previous post with another answer in case you missed it.)

OK, so what is [itex]\displaystyle\int u \,du \,?[/itex]

 

[Celadon]

(u²)/2? :x

 

[SammyS]

(u²)/2? :x

What does the ":x" mean ?

 

[Celadon]

What does the ":x" mean ?

Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.
Okay so, I tried it again like this:

u=e^x
du=e^xdx
dx=du(1/e^x)

∫u²dx-2x+∫u^-2dx = ∫u²(1/u)du-2x+∫u^-2(1/u)du
=∫udu - 2x + ∫u^-3du
=u²/2 - 2x + (u^-2/-2) + C
=u²/2 - 2x - 1/(2u²) + C
=e^2x/2 - 2x - 1/(2e^2x) + C

 

[SammyS]

Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.
Okay so, I tried it again like this:

u=e^x
du=e^xdx
dx=du(1/e^x)

∫u²dx-2x+∫u^-2dx = ∫u²(1/u)du-2x+∫u^-2(1/u)du
=∫udu - 2x + ∫u^-3du
=u²/2 - 2x + (u^-2/-2) + C
=u²/2 - 2x - 1/(2u²) + C
=e^2x/2 - 2x - 1/(2e^2x) + C

Correct.

It would likely be expressed in a manner closer to the given problem.

e^2x/2 - 2x - e^-2x/2 + C

or

(e^2x - e^-2x)/2 - 2x + C

 

[Celadon]

Correct.

It would likely be expressed in a manner closer to the given problem.

e^2x/2 - 2x - e^-2x/2 + C

or

(e^2x - e^-2x)/2 - 2x + C

Thank you very much for all your help! I can finally move on and not worry about this one question anymore. The sense of relief from when you solve a math problem.. words can't describe it.

I really do appreciate your help. :D

 

[BruceW]

just to add one last thing: I think the substitution that was hinted was the hyperbolic function, because that makes the integral nice and simple. But your way is good as well.

Edit: actually, its not much more simple. It just depends if you are more comfortable with sinusoids or exponentials.

 

[Mark44]

(u²)/2? :x

What does the ":x" mean ?

Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.

Celadon, that's not a good idea in a mathematics context, especially when x already has a meaning. You don't want to put anything in that prevents someone from understanding what you're trying to say.