Thx for this, but this is common practise
And i dont think you mean the integration of sin3x, but rather the integration of a trig function that is powered to a certain degree...
i found this for help cos i was having trouble and thought it may help all of you too.
Consider the expression sin3x. We will use the addition formulae and double angle formulae towrite this in a different form using only terms involving sinx and its powers.
We begin by thinking of 3x as 2x + x and then using an addition formula:
sin3x = sin(2x + x)= sin2xcosx + cos2xsinx
using the first addition formula=
(2sinxcosx)cosx + (1 − 2sin2x)sinx
using the double angle formula
cos2x = 1 − 2sin2x= 2sinxcos2x + sinx − 2sin3x= 2sinx(1 − sin2x) + sinx − 2sin3x
from the identity cos2x + sin2x = 1= 2sinx − 2sin3x + sinx − 2sin3x= 3sinx − 4sin3x
We have derived another identity
sin3x = 3sinx − 4sin3x
Note that by using these formulae we have written sin3x in terms of sinx (and its powers). You could carry out a similar exercise to write cos3x in terms of cosx.
it came from this website
http://72.14.203.104/search?q=cache:...u&ct=clnk&cd=1
Thx for this, but this is common practise
And i dont think you mean the integration of sin3x, but rather the integration of a trig function that is powered to a certain degree...
k thanx lol. can u please do me a huge favour and tell me the answer you get for
the integration of sin^3 x between pi/6 and zero. i keep getting the answer around the wrong way?!?
I got (3.sqrt3)/8 - 2/3
i did too, but the answer is around the other way ie 2/3 - (3.sqrt3)/8
it is question 7b in the maths in focus text book
You stated the limits as between pi/6 and zero, which implies pi/6 --> 0, and the answer is what Riviet got.
If however the limits are 0-->pi/6, then the answer is the other way around.
Because I = ∫(a-->b) f(x) dx = -∫(b-->a) f(x) dx
Hope this helps.
sorry, i mean question 7b out of exercise 5.11
yeah thanx
Icycloud read my mind. XD
Basically, if it was:
pi/6
∫ then we get our answer but
0
0
-∫ it would be reversed, by icycloud's reasoning.
pi/6
Anyway, it's all cool now.
Last edited by Riviet; 21 Mar 2006 at 5:53 AM.
actually we are all wrong. it was a simple error, when you integrate sinx, you get -cosx, thats where i went wrong, i integrated sinx to equal cosx.
Hang on a minute...so you actually do mean the integration of sin3x, and not sin^3 x
Why are you going through all that mumbo jumbo for?
The integration of sin3x is just -(cos3x)/3
And if you were to find the integration of sin^3 x, it will be easier to use substitution
sin^3(x) = (1 - cos^2 (x) ) . sinx
Then let u = cos(x), du = -sin(x)dx
Then you get:
-∫1-u^2 du
-u + (u^3)/3
Sub u back in, then sub in original borders...or change the borders first, then sub new borders into that equation
Last edited by _ShiFTy_; 21 Mar 2006 at 4:47 PM.
no, i do mean sin^3x. its just that when i integrated sin after rearanging sin^3x to equal 1/4(3sinx-sin3x) i got cos instead of negative cos for the integration of sin. i didnt want to use a substitution either because that still means that you have to expand the answer, and i find this way to be much easier.
anyways, i'm past that question, but have another one if you wouldnt mind helping?!?
The world is waiting for your question
thanx. integration of cos^3x. im not sure where to start with it
(Cos^{3}@) can be integrated by using the complex identity Cos@ = e^{@} + [e^{-i@}][/ 2Originally Posted by Newberry
where e^{i@}=Cos@ + iSin@
let z = e^{i@}
1/z = e^{-i@}
((z + (1/z))/2)^{3} = Cos^{3}@
Go from there
i dont do extension 2, and we havent done "e" yet either - any other way to do it?
Oh Sorry..Umm I'm not sure of another way...Originally Posted by Newberry
'tis ok, thanx anyway
Use subsitution, its a lot easier than you think
cos^{3}@ = (1 - sin^{2}@) x cos@
Therefore the:
∫cos^{3}@ d@
=∫ (1 - sin^{2}@) x cos@ d@
Let u = sin@
so du = cos@.d@
Substituting these in, you get the
∫ (1 - u^{2} )du ....... *now this is very easy to integrate
u - u^{3}/3
By substituting "u" back in, we get
sin@ - (sin^{3}@)/3
^{}
^{}
there is another way...
realise that cos3x = cos(2x+x) = cos2xcosx - sin2xsinx
= (2cos^2x - 1)cosx - (2sinxcosx)sinx
= 2cos^3x - cosx - 2(1-cos^2x)(cosx)
=4cos^3x - 3cosx
SO...since cos3x = 4cos^3x - 3cosx
THEN ==> cos^3x = 1/4[cos3x +3cosx], which you can integrate.
B Comm/LLB @ USyd (V)
Yeah, I feel old...as an undergrad. lol.
thanks, that was what i was trying to do, but i wasnt sure if you could multiply 2cos^2x by cosx, but apparently you can. thnx again
a pleasure
B Comm/LLB @ USyd (V)
Yeah, I feel old...as an undergrad. lol.
integral of sinx x is cosxOriginally Posted by Newberry
but integral of cos x is -sinx
It's the other way around YBK.Originally Posted by YBK
OOPS!Originally Posted by icycloud
We only did differentiation at school... that's what I was thinking of
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