Integration By Parts: Solving \int e^{2x}sin(e^x)dx

In summary, the homework statement is a problem involving integration by parts. The attempt at a solution involves using integration by parts on the "new" problem. The solution checks out and is correct.
  • #1
Sparky_
227
5

Homework Statement



[tex] \int e^{2x}sin(e^x)dx [/tex]

Homework Equations



Can I make the substitution:

[tex] w = e^x ; dw = e^x dx [/tex]

Making a "new / simpler" problem:

[tex] \int w sin(w) [/tex]


The Attempt at a Solution



Using integration by parts on the "new" problem:

[tex] u = w ; dw = du [/tex]
[tex] dv = sin (w) ; v = -cos(w)[/tex]

[tex] \int w sin(w) dw = -w cos(w) +\int cos(w) dw [/tex]

[tex]= -w cos(w) + sin(w)[/tex]

[tex]= -e^xcos(e^x) + sin(e^x) [/tex]

is this correct?

This integral is part of a larger problem and this term should "go away" supposedly.

If this is correct (this solution does not simplify to 0), then I will need to post the larger problem -
Thanks for the help
-Sparky
 
Last edited:
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  • #2
Your problem isn't really that clear to me, is this correct ...

[tex]\int e^{2x}\sin(e^x)dx[/tex]
 
  • #3
Yes - sorry I was still making my problem presentable when you replied.

It's impressive how quickly you replied.
 
  • #4
Sparky_ said:
It's impressive how quickly you replied.
:-]]]

Your initial sub. is perfect so no problem there.

I also get your final answer except with +C at the end which should always be included with indefinite integrals.
 
Last edited:
  • #5
Well it checks out and is correct. If you want to know if your Integration is correct, just take the derivative of your answer.
 
  • #6
Sometime soon, I'll try to post the entire problem - it's a differential equation from a book.

I'm not in school but I am trying to brush back up. I have the answer to it - 3 terms summed. I have 4 terms - 3 agree with the 3 - I have an extra.

I'll try to post it perhaps tomorrow. - It's on about 5-6 pages of paper.

I'll condense as appropriate.

thanks for the help.
 
  • #7
Gotcha, I'm subscribed.
 
Last edited:
  • #8

1. What is integration by parts?

Integration by parts is a technique used in calculus to solve integrals that involve products of functions. It involves breaking down the integral into smaller, more manageable parts and using the product rule to solve it.

2. How do I know when to use integration by parts?

Integration by parts is typically used when the integrand contains a product of two functions, one of which becomes simpler when differentiated and the other becomes simpler when integrated.

3. What is the formula for integration by parts?

The formula for integration by parts is ∫ u dv = uv - ∫ v du, where u and v are the functions being integrated and differentiated, respectively, and du and dv are their differentials.

4. How do I apply integration by parts to solve \int e^{2x}sin(e^x)dx?

To solve this integral, you would first identify u and dv by using the product rule. In this case, u = sin(e^x) and dv = e^{2x} dx. Then, you would find du and v by differentiating and integrating u and dv respectively. Finally, you would plug these values into the integration by parts formula and solve for the integral.

5. Can I use integration by parts for definite integrals?

Yes, integration by parts can be used to solve definite integrals as well. In this case, you would apply the integration by parts formula and then use the limits of integration to evaluate the integral.

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