Turn on thread page Beta

    Being stupid........how would you integrate cos^3 x ?

    rep for help
    cheers
    rosie
    Offline

    0
    ReputationRep:
    (Original post by crana)
    Being stupid........how would you integrate cos^3 x ?

    rep for help
    cheers
    rosie
    Can you split it up into (cosx)(cos²x), and then use a trig identity for (cos²x) and use integration by parts?

    BTW I can't be bothered doing it :rolleyes:
    Offline

    10
    ReputationRep:
    (Original post by crana)
    Being stupid........how would you integrate cos^3 x ?

    rep for help
    cheers
    rosie
    cos^3 x = cos^2 x.cos x

    Use cos^2 x = 1 - sin^2 x

    (1 - sin^2 x)cos x = cos x - sin^2 xcos x

    Integrate the second bit using a substitution, u = sin x (so du = cos x dx).

    (Original post by Pixelfairy #1)
    Can you split it up into (cosx)(cos²x), and then use a trig identity for (cos²x) and use integration by parts?

    BTW I can't be bothered doing it :rolleyes:
    cheers - my friend just suggested that too.
    i dont thinnk you actually need to integrate by parts? but i could be wrong

    rsoei
    Offline

    10
    ReputationRep:
    (Original post by crana)
    cheers - my friend just suggested that too.
    i dont thinnk you actually need to integrate by parts? but i could be wrong

    rsoei
    See above, you can use substitution.

    i used the "what would you differentiate to get this?" method

    Have hit some problems integrating sin2xsin3x though... have got a hideous expression most probably wrong......ideas?
    Offline

    15
    ReputationRep:
    (Original post by crana)
    i used the "what would you differentiate to get this?" method

    Have hit some problems integrating sin2xsin3x though... have got a hideous expression most probably wrong......ideas?
    2sinAsinB = cos(A-B) - cos(A+B)

    sin2xsin3x = [cos(2x-3x) -cos(2x+3x)]/2

    sin2xsin3x = [cos(-x) - cos(5x)]/2

    Integrate to get

    [-sin-x - sin5x/5]/2
    Offline

    1
    ReputationRep:
    (Original post by crana)
    i used the "what would you differentiate to get this?" method

    Have hit some problems integrating sin2xsin3x though... have got a hideous expression most probably wrong......ideas?
    You can use the factor formulae
    cos a - cos b = -2sin((a+b)/2)sin((a-b)/2)

    Make a = 5x, b = x.
    Offline

    0
    ReputationRep:
    (Original post by crana)
    i dont thinnk you actually need to integrate by parts? but i could be wrong.
    Just wondering, couldnt we just do (cosx)^3, then do simple chain rule?

    Bah this is all beyond me anyway.
    Offline

    15
    ReputationRep:
    (Original post by Spy_Lord)
    Just wondering, couldnt we just do (cosx)^3, then do simple chain rule?

    Bah this is all beyond me anyway.
    Chain rule is differentiation, not integration
    Offline

    0
    ReputationRep:
    (Original post by Spy_Lord)
    Just wondering, couldnt we just do (cosx)^3, then do simple chain rule?

    Bah this is all beyond me anyway.
    chain rule is for defferinating not intergration, so no. AFAIK
    Offline

    0
    ReputationRep:
    (Original post by chud)
    chain rule is for defferinating not intergration, so no. AFAIK
    I mean doing the opposite of chain rule . . strange because I recall having done it frequently. What I mean is add 1 to the indice, divide by the indice multiplied by the derivative of what's in the brackets:

    Int(cosx)^3
    =[(cosx)^4]/(4sinx) + c

    I realise though I could very well be wrong as I am only doing Add Maths right now in Year 11, which is a notch below full A Level.

    Actually, after reading some of the replies, I think perhaps opposite of chain rule cannot be done because you want expression without c. Oh well :rolleyes: .
    Offline

    14
    ReputationRep:
    (Original post by Spy_Lord)
    I mean doing the opposite of chain rule . . strange because I recall having done it frequently. What I mean is add 1 to the indice, divide by the indice multiplied by the derivative of what's in the brackets:

    Int(cosx)^3
    =[(cosx)^4]/(4sinx) + c

    I realise though I could very well be wrong as I am only doing Add Maths right now in Year 11, which is a notch below full A Level.

    Actually, after reading some of the replies, I think perhaps opposite of chain rule cannot be done because you want expression without c. Oh well :rolleyes: .
    You can't do that (but I can't remember why). aybe someone with a bigger brain will explain.

    MB
    Offline

    1
    ReputationRep:
    (Original post by musicboy)
    You can't do that (but I can't remember why). aybe someone with a bigger brain will explain.

    MB
    Only applicable if the thing in the brackets is a linear and simple term, eg 4x, you can't apply with trig or polynomials greater than 1.
    Offline

    0
    ReputationRep:
    (Original post by Bhaal85)
    Only applicable if the thing in the brackets is a linear and simple term, eg 4x, you can't apply with trig or polynomials greater than 1.
    Hey thanks for that, I had no idea. No wonder I couldn't recall doing such a question with trig.
    Offline

    14
    ReputationRep:
    (Original post by Bhaal85)
    Only applicable if the thing in the brackets is a linear and simple term, eg 4x, you can't apply with trig or polynomials greater than 1.
    I know the limitations, I was requesting an explanation of them.

    MB
    Offline

    0
    ReputationRep:
    Using my very limited Mathematic knowledge, I would just integrate cosx three times and multiply!
    Offline

    1
    ReputationRep:
    (Original post by tomhitchings)
    Using my very limited Mathematic knowledge, I would just integrate cosx three times and multiply!
    lol. But its a product of a product of a product.
    Offline

    0
    ReputationRep:
    u=sinx

    du/dx=cos x

    du/cosx=dx

    int cos^3(x) dx = int cos^2(x) du = int 1 - sin^2(x) du
    =int 1 - u^2 du

    = u - u^3/3

    = sinx -(sin^3(x)/3)
 
 
 

University open days

  • University of Warwick
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • University of Sheffield
    Undergraduate Open Days Undergraduate
    Sat, 20 Oct '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 20 Oct '18
Poll
Who is most responsible for your success at university

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.