Integration by Parts (1 Viewer)

[tommykins]

This is driving me nuts, I've done it 2-3 times and always get a variation of the answer.

Answer - [e^(pi/2) - 1]/2

int (pi/2 -> 0) (e^x).(cosx) dx

Integration by parts twice.

EDIT - Gah fuck, nevermind. Got it.

 

[Pwnage101]

Its not too bad, but does take a little thinking:

Let: int (pi/2 -> 0) (e^x).(cosx) dx = I
by parts: u=(e^x) v'= (cosx)
u'=(e^x) v= (sinx)

that is I = [(e^x)(sinx)] from (pi/2 -> 0) - int (pi/2 -> 0) (e^x).(sinx) dx.......(1)

now, consider int (pi/2 -> 0) (e^x).(sinx) dx
by parts: u=(e^x) v'=(sinx)
u'=(e^x) v=(-cosx)

that is, int (pi/2 -> 0) (e^x).(sinx) dx = [(e^x)(-cosx)] from (pi/2 -> 0) - int (pi/2 -> 0) (e^x).(-cosx) dx
= [(e^x)(-cosx)] from (pi/2 -> 0) + int (pi/2 -> 0) (e^x).(cosx) dx

but, we defined 'int (pi/2 -> 0) (e^x).(cosx) dx' to equal I

= [(e^x)(-cosx)] from (pi/2 -> 0) + I

sub this bak into (1):
I = [(e^x)(sinx)] from (pi/2 -> 0) - int (pi/2 -> 0) (e^x).(sinx) dx.......(1)

I = [(e^x)(sinx)] from (pi/2 -> 0) - {[(e^x)(-cosx)] from (pi/2 -> 0) + I}
= [(e^x)(sinx)] from (pi/2 -> 0) + [(e^x)(cosx)] from (pi/2 -> 0) - I

TAke the I to the other side, ie:

2I = [(e^x)(sinx)] from (pi/2 -> 0) + [(e^x)(cosx)] from (pi/2 -> 0)
= (e^(pi/2)) - 1

that is,

I = {e^(pi/2) - 1}/2

Gotta watch all the negative and positive signs in this integration, easy to make a silly mistake

NB, square brakets indicate the integral has been found, but needs to be evaluated between the 2 limits

 

[midifile]

Let I = int (pi/2-->0) (e^x).(cosx)dx
Let u = e^x
du = e^x
v=sinx
dv = cosx
Therefore I = [sinx e^x](pi/2-->0) - int(pi/2-->0) e^xsinx dx
For the integral, let u = e^x
du = e^x
v=-cosx
dv = sinx
I= e^(pi/2)- 0 + [e^xcosx](pi/20) – int(pi/2-->0) e^xcosx dx
I= e^(pi/2) + 0 – e^0 – I
2I = e^(pi/2) – 1
Therefore I = (e^(pi/2) – 1)/2

I hope there are no mistakes in there. I typed it straight in to the computer

 

[conics2008]

hey let me try out my new scanner xD

im sorry... i know u have done it.. im assuming your mistake was gettin the 2In

 

[conics2008]

one thing is it a must to identify the v u crap.. because i dont use them.. i just apply them as i go along.. which makes the working out much shorter and efficent.

 

[Iruka]

Well, it is outside the syllabus, but you could use complex numbers. Then you don't have to use integration by parts at all.

cos(x)= 1/2(e^(ix) + e^(-ix))

You just treat i like an ordinary constant, and it all works out in the end. (You will eventually get a real valued function out of doing the integration.)

EDIT: you will probably also need to know that sin(x) = 1/2i (e^(ix) - e^(-ix))

 

[Pwnage101]

one thing is it a must to identify the v u crap.. because i dont use them.. i just apply them as i go along.. which makes the working out much shorter and efficent.

i believe it is not necessary, helps me think it out and see out any mistakes, but yeh when ur dealing with e^x and cos/sin x its pretty easy to integrate by parts

 

[Pwnage101]

btw, nice to see the MX2 community is so helpful - if this were just about any other subject forum , i highly doubt one would have four excellent solutions posted up within 20 minutes

 

[conics2008]

^^^ i didn't get what u mean...

eg do i have to write what my v and v' etc etc is.. in order to get awarded my neccsary marks... because it seems like a waste of time... ????

 

[Pwnage101]

^^^ i didn't get what u mean...

eg do i have to write what my v and v' etc etc is.. in order to get awarded my neccsary marks... because it seems like a waste of time... ????

what i meant is what i wrote , from what i have heard from my teacher, it is NOT NECESSARY to do that to gain the marks, ie NO, u DO NOT HAVE TO write v',v ',u, u', and all (to spell it out for u even more simply)

but thats what ive heard, and it may not be what HSC markers expect - "all we know is what we're told, and for all we know that isn't even true" ..."for all anyone knows, nothing is...Audiences act on assumptions..." lol i cant believe i could remember a quote from Ros and Guil are dead!!!!!

 

[conics2008]

relax its on maths.. lol..... hahahaah ahahahah at you !

 

[tommykins]

回复: Re: Integration by Parts

hey let me try out my new scanner xD

im sorry... i know u have done it.. im assuming your mistake was gettin the 2In

Nah, I had everything right but forgot during the -1 instead I got +1, skipped too many steps.

Thanks for all your inputs/help =D

Random Q -

Is the eccentricity of xy = 8c^2/9 = sqrt2?

 

[tacogym27101990]

yeah all rectangular hyperbolas have eccentricity of rt2

 

[DennyCrane]

tommykins, do yo u have Mrs Rankin or Mr. Donnolly for 4unit?

 

[tommykins]

回复: Re: Integration by Parts

tommykins, do yo u have Mrs Rankin or Mr. Donnolly for 4unit?

Hughes and Turbett are taking 4u this year.

I have Hughes.

 

[melonkitten]

god, i hate it when you try by parts for agessssssssssssssssssssss,, and it turns out you've done it the wrong way round

 

[Slidey]

god, i hate it when you try by parts for agessssssssssssssssssssss,, and it turns out you've done it the wrong way round

Haha. It's all experience I guess.

 

[Iruka]

Hehe.

What I really love about integration by parts is when you have to do it twice, and you get the second one the wrong way round. You end up with something like "0=0."

Very true. And completely unenlightening.