Calculus

Integral of sin2xcos3x using Integration by Parts

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  1. so, one way is to let
    u = sin2x, du = 2 cos2x dx
    dv = cos3x, v = 1/3 sin3x
    Now we have
    I = ∫sin2x cos3x dx = ∫ u dv
    = uv - ∫ v du
    = 1/3 sin2x sin3x - 2/3 ∫cos2x sin3x dx
    For the second integral, let
    u = cos2x, du = -2sin2x dx
    dv = sin3x dx, v = -1/3 cos3x
    ∫cos2x sin3x dx = -1/3 cos2x cos3x - 2/3 ∫ sin2x cos3x dx
    Putting all that together, we end up with
    I = 1/3 sin2x sin3x - 2/3 (-1/3 cos2x cos3x - 2/3 ∫ sin2x cos3x dx)
    I = 1/3 sin2x sin3x + 2/9 cos2x cos3x + 4/9 I
    5/9 I = 1/3 sin2x sin3x + 2/9 cos2x cos3x
    I = 3/5 sin2x sin3x + 2/5 cos2x cos3x
    That can be massaged in various ways, one of which is
    1/2 cosx - 1/10 cos5x

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    oobleck
  2. I really really like the algorithm for integration by parts shown in
    https://www.youtube.com/watch?v=2I-_SV8cwsw

    where "blackpenredpen" shows the DI method and goes through the 3-Stop method.

    -----D ------- I , D for differentiate, I for integrate
    +| sin2x .... cos3x
    -| 2cos2x .. (1/3)sin3x
    +| (-4sin2x) .. (-1/9)cos3x <--- This product is a multiple of the original, so we stop
    -| ....

    ∫sin2x cos3x dx = (1/3)sin2xsin3x + (2/9)cos2xcos3x + (4/9)∫sin2x cos3x dx
    (5/9)∫sin2x cos3x dx = (1/3)sin2xsin3x + (2/9)cos2xcos3x

    ∫sin2x cos3x dx = (3/5)sin2xcos3x + (2/5)cos2xcos3x <------ oobleck's answer

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    Reiny

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