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Integral - is this correct?

  1. Oct 27, 2006 #1
    This has been posted before although I've come across it and got a differant answer from https://www.physicsforums.com/archive/index.php/t-108378.html

    x sinx cosx dx using the identity sin2x = 2sinxcosx

    u = x
    du = 1
    dv = 1/2 sin 2x
    v = -1/4 cos 2x

    x sinx cosx dx = -1/4xcos2x - Int -1/4cos2x + c
    = -1/4xcos2x + 1/8sin2x + c
    Right?

    And I need some help to work out:
    Integral of x^3cosx^2

    I'm always getting confused with powers on the trig for some reason.
    Thankyou
     
  2. jcsd
  3. Oct 27, 2006 #2

    quasar987

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    I would agree with that.

    Your second integral is also solvable by parts. The trick for choosing u and dv is you gotta choose dv such that [itex]\int dv[/itex] is feasable! Based on that, what are you gonna choose for u and dv?
     
  4. Oct 28, 2006 #3
    Well lets try...

    Int x^3cos(x^2)

    u = cos(x^2)
    du = -2x sin (x^2)
    dv = x^3
    v = (x^4)/4

    Int x^3cos(x^2) = (x^4cos(x^2))/4 + 1/2 Int x^5 sin (x^2)

    u = sin(x^2)
    du = 2x cos(x^2)
    dv = x^5
    v = (x^6)/6

    Int x^5 sin (x^2) = (x^6sin(x^2))/6 - 1/3 Int x^7 cos (x^2)

    Hmmm... Doesn't seem to be working out that way... Unless I look at the pattern and find...

    Int x^3cos(x^2) = (x^4cos(x^2))/4 + 1/2((x^6sin(x^2))/6) - 1/3((x^8cos(x^2))/8) + 1/4((x^10sin(x^2))/10).......

    Although thats just a guess...
    Damn year 12...
     
  5. Oct 28, 2006 #4
    Some how the answer comes out to be:

    1/2 (x^2sinx^2 + cosx^2) + c

    and I hope I didn't make a mistake in my calculations up there...
     
  6. Oct 28, 2006 #5

    quasar987

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    Alright, I guess my "trick for choosing u and dv" isn't so good afterall because what you took as dv has an easily found primitive but you got nowhere with it.

    What other choice of u and dv can you make? There aren't an infinity, there are only two:

    - There's u=x, dv=x²cos(x²). You can try doing it with this if you know the primitive of wcos(w). I haven't tried it so I don't know if it will work.

    - There's u=x², dv=xcos(x²). Can you see that this choice of dv makes it perfectly fit for an integration by change of variable?


    You might want to look at the ILATE rule on wiki that gives a general rule of thumb for what to take as u and dv. There is a remark at the end however that warn the reader that there are exceptions to the ILATE rule. This problem of yours is one of them.

    http://en.wikipedia.org/wiki/Integration_by_parts#The_ILATE_rule
     
    Last edited: Oct 28, 2006
  7. Oct 28, 2006 #6
    Let [tex] u = x^{2} [/tex] and [tex] dv = x\cos(x^{2}) dx [/tex]


    Then [tex] du = 2x dx [/tex] and [tex] v = \frac{1}{2} \sin(x^{2}) [/tex]

    Then just apply the formula from here.

    [tex] \int udv = uv - \int vdu [/tex]
     
    Last edited: Oct 28, 2006
  8. Oct 28, 2006 #7
    Didn't know you could just move over the x. Sweet.
    Thankyou both quasar987 and courtrigrad
     
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