I'm trying to integrate this function using integration by parts, but I get stuck after a while.

The function is

I start by using the formula

So taking , then

Now what do I do?

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- August 18th 2010, 05:58 PMexothesisIntegration by parts
I'm trying to integrate this function using integration by parts, but I get stuck after a while.

The function is

I start by using the formula

So taking , then

Now what do I do? - August 18th 2010, 06:05 PMProve It
No need for integration by parts, use trigonometric identities.

. - August 18th 2010, 06:10 PMexothesis
The question states to use integration by parts, I assume there would be a way?

- August 18th 2010, 06:30 PMpickslides
You may need to integrate by parts twice here.

and

Now

What do you get? - August 20th 2010, 05:19 AMHallsofIvy
- August 22nd 2010, 10:30 PMexothesis
I end up with 0? I don't think I'm doing it right. This is what I do:

But then all together that means:

What am I doing wrong here? My CAS calculator says the answer should be

- August 22nd 2010, 11:18 PMBayernMunich
Solving this one with integration by parts is a time's wasting.

Use this:

- August 23rd 2010, 03:30 AMDefunkt
- August 23rd 2010, 03:46 AMtom@ballooncalculus
http://www.ballooncalculus.org/forum...lllpartslllone

Ok, it's a video, but my link is to a .wmv file and I need to do it differently, clearly. Will fix. (Edit: fixed, but you have to follow the link to see the video - was hoping it would project right here!)

Anyway, just in case a picture helps...

http://www.ballooncalculus.org/draw/parts/one.png

... where (key in spoiler) ...

__Spoiler__:

________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - August 23rd 2010, 04:34 AMProve It
If you MUST use integration by parts, then rewrite your integral as

Now apply integration by parts with so and so .

The integral becomes

So this gives

.

Apply integration by parts once more, then you will have your answer. - August 23rd 2010, 04:45 AMtom@ballooncalculus
- August 23rd 2010, 07:19 PMexothesis
Thanks for all the help. I should have no trouble solving it using the methods described above.

- August 23rd 2010, 09:41 PMSoroban
Hello, exothesis!

The formula I use has this form: .

Quote:

Integrate by parts: .

By parts: .

And we have: .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

So we have: .

. . . . . . . . .

. . . . . . . . .

Therefore: . .