Math help

[RESmonkey]

Integrate this, please: (e^(-x))cosx dx

I tried using integration by parts, aka uv - (integration of v)du but process keeps repeating itself (from cos to sine, and then back agian, etc.)

 

[KLin]

42

 

[compuwiz1]

<canned response> Do your own damn homework! </canned response>

 

[RESmonkey]

...

 

[Parasitic]

integral of exp(ax) cos(bx) dx = exp(ax) *(a cos bx + b sin bx)/(a^2+b^2)
a= -1
b= 1
so
the integral is exp(-x)*(-cos x+ sin x)/2.

Why? Because I said so. Also it's equation #332 on p.175 of Mathematics Handbook for Science and Engineering, 5th ed. by Rade and Westergren, Springer-Verlag publishing.

Moral of this lesson: get yourself a math handbook.

 

[JohnCU]

of course the integration keeps repeating. do it until you get back to the original integral, then set it equal to whatever you have so far and solve that way.

 

[RESmonkey]

^Whaa?!??!!?

Did you just pull some Calculus 7 and a half on me?

Sorry, But I <emphasized-italics>fail[/]</emphasized-italics> to understand that

 

[potato28]

Originally posted by: KLin
42

QFT!

 

[RESmonkey]

Wait. So I set the orig. thing equal to itself? Huh?

 

[Parasitic]

Originally posted by: RESmonkey
^Whaa?!??!!?

Did you just pull some Calculus 7 and a half on me?

Sorry, But I fail to understand that

Didn't I just post the solution up ahead?
Use your integral table. There's one in every Calculus book. Maybe not as extensive but can get you a head start.

BTW you have much to learn about italising, young Skywalker.

 

[JohnCU]

Originally posted by: RESmonkey
^Whaa?!??!!?

Did you just pull some Calculus 7 and a half on me?

Sorry, But I <emphasized-italics>fail</emphasized-italics> to understand that

dude, this is calc II stuff. do an integration by parts. you get like sin and something. hold all that. do another integration by parts (on the sin integral). you get a really long string if you are keeping up with it. then you get back what you started with. subtract the new repetition of the original equation to the other side, you'll be able to pull out a constant, divide both sides and get an answer.

 

[RESmonkey]

I did that and got this:

(original integration thing) = (Original integration thing) + e^(-x)*cosx

I subtract over the original, and get :

0 = e^(-x) * cosx

What does that mean?

 

[Parasitic]

You fail.

 

[RESmonkey]

Is my last post's work right? I think I need a constant somewhere in there.

 

[JohnCU]

do this:

int(e^-x*cosxdx) = e^-x*sinx+int(e^-x*sinxdx)

=> int(e^-x*cosxdx) = e^-x*sinx+[-e^-x*cosx-int(e^-x*cosxdx)]

=> 2*int(e^-x*cosxdx) = e^-x*sinx - e^-x*cosx

=> int(e^-x*cosxdx) = (1/2)*e^-x*sinx - (1/2)e^-x*cosx

 

[Fenixgoon]

or do a laplace transform.. mmmm, laplace transform

 

[Parasitic]

Originally posted by: Fenixgoon
or do a laplace transform.. mmmm, laplace transform

Shhh you're scaring the calc newbie!
Plus that wouldn't make the problem easier.

 

[RESmonkey]

Holy shit. Thank you. I understand now...

I should really pay attention in class...

 

[Eeezee]

PI IS EXACTLY EQUAL TO THREE!

 

[Fenixgoon]

Originally posted by: Parasitic

Originally posted by: Fenixgoon
or do a laplace transform.. mmmm, laplace transform

Shhh you're scaring the calc newbie!
Plus that wouldn't make the problem easier.

trying to think.. what would that be?

(-1/s) * 1/(s^+1) ?

probably more trouble than it's worth for this problem, you're right. so much better for solving diff eq's.. or at least the "textbook" ones.

 

[chuckywang]

Originally posted by: JohnCU
of course the integration keeps repeating. do it until you get back to the original integral, then set it equal to whatever you have so far and solve that way.

Yep, pretty much.

 

[frostedflakes]

Yup, it's a recursive (I think that's what they're called) integral. You end up with some stuff minus the original integral being equal to the original integral. You then add the original integral to both sides so you have the stuff you've obtained by parts equal to 2x the original integral. Then divide both sides by two.

Sounds like you get it but if not let me know and I can try to work it out on paper and post up a scan. For me it makes stuff easier to understand when I can see the whole thing worked out and can see step by step what's going on.