Integration by Parts

[nasi112]

This forum does not allow us to reply to old post, so I will rewrote @rooney post because I have some questions.

[math]\int e^{-st}\sin t dt[/math]
I see 4 options to solve this integration

Option 1
[math]u = e^{-st}[/math][math]dv = \sin t dt[/math]
Option 2
[math]u = \sin t [/math][math]dv = e^{-st} dt[/math]
Option 3
[math]u = e^{-st}\sin t [/math][math]dv = dt[/math]
Options 4
[math]u = 1[/math][math]dv = e^{-st}\sin t dt[/math]
I saw the solution of @galactus and he chose Option 1. How did he know this option will work and other option will fail?

 

[Dr.Peterson]

This forum does not allow us to reply to old post, so I will rewrote @rooney post because I have some questions.

You are referring to

integration by parts: integral( (e^-s*t)*sin(t) dt)

I'm trying to find a Laplace transform, and getting stuck in the integration. I have: integral( (e^-s*t)*sin(t) dt) i tried using u = e^-s*t and v' = sin(t) which gave me (e^(-s*t))(-cos(t))-integral((-s*e^(-s*t))*-cos(t) dt) i feel like i am back where i started.

www.freemathhelp.com

It's not that you are not allowed to reply to an old post, but that if you want to add to such a thread, you should be asking your own question, not pretending you are talking to the OP, who may not even be alive. What you're doing here is fine, and would be fine in that thread, as far as I'm concerned.

I saw the solution of @galactus and he chose Option 1. How did he know this option will work and other option will fail?

I would guess that either (a) he tried one or more of the options, and found that that one worked, or (b) he had enough experience (from having tried various options for similar problems before) to expect that one to work.

Trying things until one of them works is a normal practice in this sort of problem. Various mnemonics, such as the recent writer suggested, are just a way to turn such experience into a formula; I prefer experience! In particular, it will be very useful for you to try all four, in order to see for yourself whether only one works, and what it takes to make it/them work.

In this case, my experience would tell me that either of your first two options is reasonable, because the u can be easily differentiated and the dy is easily integrated; but that neither results in a simpler integrand, so I would hope for the sort of thing that, indeed, happens with either.

Whenever someone asks me, "Why do they do it that way? Why not this other way?" I always say, "Try doing it the other way and see what happens. You'll either find that it doesn't matter, or understand deeply why the one way is better." Either way, you've really learned something.

 

[Krishang]

How did he know this option will work and other option will fail?

Hey There! One can think intuitively Here. Option 4 is the question itself once you apply IBP. Option 3 isn't preferable as after differentiation it will become complicated. Here, both options 1 and 2 can work here. Since both of these never become 0 after differentiation using IBP twice will loop it back to a form of

[math]\int e^{-st} dt = f(x) + g(x) ± \frac{1}{a}\int e^{-st}[/math]
You may also think of this intuitively. You should avoid the second integral to get complicated and try to choose the second term in IBP as the easier one to integrate. Also try to make the terms in similiar forms in general. However, selecting u and dv will take time and practice. There isn't any "hard and fast rule". Like ILATE is also a general rule, and you try to take dv as the one easier to integrate. But ILATE may also give you the wrong order in solving, though rarely. For example, [math]\int \frac{xe^x}{(1+x)^2} dx[/math].

I would guess that either (a) he tried one or more of the options, and found that that one worked, or (b) he had enough experience (from having tried various options for similar problems before) to expect that one to work.

Trying things until one of them works is a normal practice in this sort of problem. Various mnemonics, such as the recent writer suggested, are just a way to turn such experience into a formula; I prefer experience! In particular, it will be very useful for you to try all four, in order to see for yourself whether only one works, and what it takes to make it/them work.

In this case, my experience would tell me that either of your first two options is reasonable, because the u can be easily differentiated and the dy is easily integrated; but that neither results in a simpler integrand, so I would hope for the sort of thing that, indeed, happens with either.

I also support this. It will be nice to try in future problems different u and dv and seeing which one works and why. Though there are some general points to keep in mind, sometimes thinking out of the box can be the thing you needed to solve the problem, and that can be achieved through practice only.