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    i have a couple of questions im stuck on to do with p3

    1) how do you integrate (lny)/y ?

    2)
    how do you integrate (sin3x - cos3x)/(sin2x cos2x)

    this question is the last part of quite a long question, in the first parts i found that;
    d(secx)/dx = secx tanx
    d(cosecx)/dx = -cosecx cotx
    cosecx + secx = 2(sinx + cosx)/sin2x
    and d((sinx + cosx)/sin2x)/dx = 0.5(tanx secx - cotx cosecx)
    i figure that to answer question 2, these identities must be used somehow, but i'm not quite sure how. its only worth 3 marks so it can't be too complicated.

    thanks
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    1. See yazan's post below. [edit]

    2.
    ∫ (sin³x - cos³x)/(sin²x cos²x) dx = ∫ sin³x/(sin²x cos²x) dx - ∫ cos³x/(sin²x cos²x) dx
    = ∫ sinx/cos²x dx - ∫ cosx/sin²x dx
    = ∫ tanx secx dx - ∫ cotxcosecx dx
    = secx + cosecx + C
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    1) how do you integrate (lny)/y ?
    use u = lny
    du/dy = 1/y
    dy = y du

    int ln(y) / y dy = int u/y .ydu = int u du = ½u² +C = ½(lny)² + C
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    (Original post by gemm)
    i have a couple of questions im stuck on to do with p3

    1) how do you integrate (lny)/y ?

    2)
    how do you integrate (sin3x - cos3x)/(sin2x cos2x)

    this question is the last part of quite a long question, in the first parts i found that;
    d(secx)/dx = secx tanx
    d(cosecx)/dx = -cosecx cotx
    cosecx + secx = 2(sinx + cosx)/sin2x
    and d((sinx + cosx)/sin2x)/dx = 0.5(tanx secx - cotx cosecx)
    i figure that to answer question 2, these identities must be used somehow, but i'm not quite sure how. its only worth 3 marks so it can't be too complicated.

    thanks
    for part 2:

    (sin3x - cos3x)/(sin2x cos2x)

    this is equal to: (sin3x)/(sin2x cos2x) - (cos3x)/(sin2x cos2x)

    this simplifies to sinx/cos2x - cosx/sin2x
    which is secx tanx - cosecx cotx

    INT (secx tanx - cosecx cotx) dx
    from your info
    = secx + cosecx + C
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    cheers,
 
 
 
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