Integration of sin3x (1 Viewer)

[Newberry]

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...oubleangle.pdf+sin3x&hl=en&gl=au&ct=clnk&cd=1

 

[_ShiFTy_]

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...

 

[Newberry]

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?!?

 

[Riviet]

I got (3.sqrt3)/8 - 2/3

 

[Newberry]

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

 

[icycloud]

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.

 

[Newberry]

sorry, i mean question 7b out of exercise 5.11

 

[Newberry]

yeah thanx

 

[Riviet]

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.

 

[Newberry]

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.

 

[_ShiFTy_]

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

 

[Newberry]

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?!?

 

[_ShiFTy_]

The world is waiting for your question

 

[Newberry]

thanx. integration of cos^3x. im not sure where to start with it

 

[rama_v]

thanx. integration of cos^3x. im not sure where to start with it

(Cos³@) can be integrated by using the complex identity Cos@ = e^@ + [e^-i@][/ 2

where e^i@=Cos@ + iSin@

let z = e^i@
1/z = e^-i@

((z + (1/z))/2)³ = Cos³@
Go from there

 

[Newberry]

i dont do extension 2, and we havent done "e" yet either - any other way to do it?

 

[rama_v]

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...

 

[Newberry]

'tis ok, thanx anyway

 

[_ShiFTy_]

Use subsitution, its a lot easier than you think

cos³@ = (1 - sin²@) x cos@

Therefore the:
∫cos³@ d@
=∫ (1 - sin²@) x cos@ d@

Let u = sin@
so du = cos@.d@
Substituting these in, you get the

∫ (1 - u² )du ....... *now this is very easy to integrate
u - u³/3

By substituting "u" back in, we get
sin@ - (sin³@)/3



 

[Mountain.Dew]

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.