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Integral - is this correct?

  1. Jan 27, 2006 #1
    I have been asked to find the integral (i) sinx cox dx and the integral
    (ii) x sinx cosx dx using the identity sin2x = 2sinxcosx

    My work...

    (i) sinx cox dx

    = 1/2 integral of 2 sinx cos dx

    = 1/2 integral of sin 2x dx

    = 1/2 cos 2x + C is this correct?

    (ii) x sinx cosx dx

    not sure how to do this one ?
  2. jcsd
  3. Jan 27, 2006 #2


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    Almost, you made one mistake: you forgot a minus.
    (Be careful: for derivatives it's the other way arround)

    Have you seen integration by parts?
  4. Jan 27, 2006 #3


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  5. Jan 27, 2006 #4
  6. Jan 27, 2006 #5


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    Didn't you forget an integral when you applied integration by parts?
    It is [itex]\int {udv = uv - } \int {vdu} [/itex] and you left out that last integral I believe...
  7. Jan 27, 2006 #6
    I don't think so no because from the previous example we get the integral of sin x cos x which I use at the end. So I think I am correct no ? :confused:
  8. Jan 27, 2006 #7


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    With v = -1/2 cos(2x) and du = dx, the formula gives a new integral at the end and not just "-1/2 cos(2x)". Applying integration by parts would give:

    [tex]\int {x\sin x\cos xdx} = - \frac{1}{2}x\cos \left( {2x} \right) - \int { - \frac{1}{2}\cos \left( {2x} \right)dx} = - \frac{1}{2}x\cos \left( {2x} \right) + \frac{1}{2}\int {\cos \left( {2x} \right)dx}[/tex]

    You forgot that last integral and just added "-1/2 cos(2x)", you see?
  9. Jan 27, 2006 #8
    must I have to do another integration by part on 1/2 integral of cos 2x
  10. Jan 27, 2006 #9


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    Well, you must do one more integration but the total process is called 'integration by parts' (which you only have to do once). Have you already covered this technique? Basically, it comes down to the formula I gave two posts back.
  11. Jan 27, 2006 #10
    from here I do:

    u = 2x
    du = 2

    = 1/2 integral of cos u du
    = 1 / 2 sin 2x + C

    so the whole thing gives - 1/2 x cos 2x + 1/2 sin 2x + C is this correct?
  12. Jan 27, 2006 #11


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    Almost, there was already a factor 1/2 before the integral and because of your substitution, there should be another one so that would give 1/4 sin(2x) for that last part.

    Then it should be correct, if you don't forget the constant of integration.
    If you want to check it yourself: find its derivative and see if you get your initial function again :smile:
  13. Jan 27, 2006 #12
    Thanks TD you are a star!
  14. Jan 27, 2006 #13


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    You're welcome :smile:
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