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    With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...


    prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c


    anyone...?
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    (Original post by scottyoneill)
    With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...


    prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c


    anyone...?
    dy/dx= 1 and ∫cos2x =sin2x/2

    xsin2x/ 2 - ∫ sin2x/2

    xsin2x/2 - (- cos 2x /4)

    xsin2x/2 + cos2x /4

    sin2A=2sinAcosA and cos2x =1 -sin^2x

    2xsinxcosx + 1 - sin^2x /4

    then im stuck where did i go wrong
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    Just use integration by parts with u=x and v´=Cos 2x
    ie u´=1 and v = 1/2 Sin 2x

    therefore

    I xCos2x dx = x . 1/2 Sin 2x - I I/2 Sin 2x dx

    = 1/2 x 2 Sin x . Cos x + 1/4 Cos 2x + c

    = 1/2 x 2 Sin x.Cos x + 1/4 (1 - 2 Sin^2 x) +c

    = 1/2 x 2 Sin x.Cos x + 1/4 - 1/2 Sin^2 x +c

    = 1/2 x 2 Sin x.Cos x - 1/2 Sin^2 x + k

    = 1/2(sinx(2xcosx-sinx) + k
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    (Original post by scottyoneill)
    With the exam on wednesday, i thought i'd start this thread for us test each other with questions to see how ready we are! I'll start the ball rolling with a question im having trouble with...


    prove that the integral of xcos2x can be expressed as 1/2(sinx(2xcosx-sinx) + c


    anyone...?
    ∫xcos2x dx

    Using integration by parts I got:

    (xsin2x)/2 + (cos2x)/4

    Cross multiplied everything to obtain:

    (4xsin2x+2cos2x)/8

    then expanded the sin2x and cos2x to get:

    (2xsinxcosx)/2 - (1-2sin^2x)/4

    [8xsinxcosx-(2-4sin^2x)]/8

    (2xsinxcosx-sin^2x)/2
    =[sinx(2xcosx-sinx)]/2
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    here's a quickie

    ∫ √(1-x^2) dx

    .. actually this may be more complicated than I thought...
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    let x = sin u etc
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    (Original post by Mrm.)
    let x = sin u etc
    The is the answer just x??? Seems very unlikely - but that's what i got!!!
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    (Original post by kimoni)
    here's a quickie

    ∫ √(1-x^2) dx

    .. actually this may be more complicated than I thought...
    -3x(1-x^2)^3/2 + c
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    (Original post by Mrm.)
    let x = sin u etc
    yeah, that's what I thought, and I get an answer, but go to http://www.calc101.com/webMathematica/integrals.jsp and type in sqrt[1-x^2] you get something completely different.. or is it just me?

    ed: actually.. maybe not completely different
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    (Original post by Silly Sally)
    The is the answer just x??? Seems very unlikely - but that's what i got!!!
    Ok then!!! Got it completely wrong!!! can someone show me how to do this question please!!!
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    i dont have a password
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    (Original post by Mrm.)
    i dont have a password
    neither do I, but it still shows you the answer

    -can anyone offer a definitive proof for this question?
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    On the Edexcel board its it always presumed that x=sinu?
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    i get something very similar to the calc101 answer....just sorting out the final bits and pieces

    you do use the sub x=sin u
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    I just did it and it comes out fine.
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    (Original post by Bhaal85)
    ∫xcos2x dx

    Using integration by parts I got:

    (xsin2x)/2 + (cos2x)/4

    Cross multiplied everything to obtain:

    (4xsin2x+2cos2x)/8

    then expanded the sin2x and cos2x to get:

    (2xsinxcosx)/2 - (1-2sin^2x)/4

    [8xsinxcosx-(2-4sin^2x)]/8

    (2xsinxcosx-sin^2x)/2
    =[sinx(2xcosx-sinx)]/2
    Sorry i don't understand :confused: - what you did for the bold bit? Can you please explain
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    Well having used x=sin u as my sub, i get

    ∫cos^2 x
    = (sin2u)/4 + u/2

    x=sin u
    u=arcsin x

    sin2(sin^-1 x)/4 + (sin^-1)/2 + C

    which is almost there.. I dont know how to do sin2(sin^-1 x)/4 though
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    (Original post by Mrm.)
    I just did it and it comes out fine.
    can you show your working then?
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    (Original post by kimoni)
    Well having used x=sin u as my sub, i get

    ∫cos^2 x
    = (sin2u)/4 + u/2

    x=sin u
    u=arcsin x

    sin2(sin^-1 x)/4 + (sin^-1)/2 + C

    which is almost there.. I dont know how to do sin2(sin^-1 x)/4 though
    Whats happened to the square root part when you integrate?
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    (Original post by Silly Sally)
    Whats happened to the square root part when you integrate?
    If you use x = sin t in (1 - x^2)^1/2, you get cos t (from the identity sin^2 t + cos^2 t = 1).
 
 
 

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