calculus

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∫ sin(x)sin(3x)dx

find the integral

  • calculus -

    one way:

    sin(3x) = sinx cos2x + cosx sin2x
    = sinx(2cos^2(x)-1) + 2cosx sinx cosx
    = cos^2(x) sinx - sinx + 2cos^2(x) sinx
    = 3cos^2(x)sinx - sinx

    ∫ = -cos^3(x) + cos(x)

    another way:

    sinx sin3x = 1/2 cos(2x) - 1/2 cos(4x)
    ∫ = 1/4 sin2x - 1/8 sin4x

    a little manipulation of that also yields

    sin^3(x) - cos(x)

    the various expressions are not identical, but differ only by a constant C

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