Complex Integration by Parts Help. Please

  • Thread starter Silmax
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  • #1
Silmax
4
0
Hi
Can anyone help me with this integration.
I will use I to symbol the integer sign

Limits between 1 and 0

Ie^-x.Sinxdx

I understand I have to integrate by parts and get the following answer, ignoring the limits for the time being.

Ie^-x.Sinx = e^-x.Cosx I -Cosx.e^-x dx

then i believe i have to integrate the second part of the sum ( I -Cosx.e^-xdx)and i get.

I -Cosx.e^-xdx= -Cosx.-e^-x- I -e^-x.Sinx

Now i am stuck.
How do I combine the two answers to get a single sum which i can then put between limits.??

Thank you.
 

Answers and Replies

  • #2
sutupidmath
1,631
4
Is this what you are trying to integrate:

[tex]I=\int_{0}^{1}e^{-x}sin(x)dx...?[/tex]

I don't see any complex integral here. I might be wrong as well, because my eyes are going...lol...

let [tex] u=e^{-x}, du=-e^{-x}dx,and,v=\int sin(x)dx=-cosx[/tex]

Integrate by parts twice, and you will get sth in terms of the original integral, and you will be fine.
 
  • #3
Moridin
688
3
The OP might be referring to the integral

[tex]\int^{1}_{0} ie^{-x} \sin (x) dx[/tex]

?
 
  • #4
Silmax
4
0
[tex]S=\int_{0}^{1}e^{-x}sinxdx.[/tex]
This is the actual sum. Now I have learned how to set it up properly.
Thanks
 
  • #5
Pere Callahan
586
1
Is this what you are trying to integrate:

[tex]I=\int_{0}^{1}e^{-x}sin(x)dx...?[/tex]

I don't see any complex integral here. I might be wrong as well, because my eyes are going...lol...

let [tex] u=e^{-x}, du=-e^{-x}dx,and,v=\int sin(x)dx=-cosx[/tex]

Integrate by parts twice, and you will get sth in terms of the original integral, and you will be fine.

There is no need for subsitution here - actually I don't see how it helps, you'd get something like [itex]\sin\log u[/itex]...not really helpful.

Just use integration by parts right away, and you'll get back to the integral you started from for which you can then solve.
 
  • #6
sutupidmath
1,631
4
There is no need for subsitution here - actually I don't see how it helps, you'd get something like [itex]\sin\log u[/itex]...not really helpful.

Just use integration by parts right away, and you'll get back to the integral you started from for which you can then solve.

I don't see myself saying to use substitution lol...!

And as long as i can see, i also suggested integration by parts, and that twice.

We might have different concepts of what substitution and integration by parts is, who knows!

OR... Probbably my eyes are really going lol...
 
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  • #7
Pere Callahan
586
1
Oh true you didn't mention substitution explicitly, my fault.

But why did write down then what du is ...hm, doesn't matter. I guess our concepts of subsitution and integration by parts are rather similar.:smile:
 
Last edited:
  • #8
sutupidmath
1,631
4
Oh true you didn't mention substitution explicitly

Neither explicitly nor implicitly!
 

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