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Calculus

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integral[13,4]sin^2(8x)cos^2(8x)dx

  • Calculus -

    sin^2(8x)cos^2(8x)
    = 1/4 (2sin(8x)cos(8x))^2
    = 1/4 sin^2(16x)
    now use integration by parts twice to evaluate.

  • Calculus -

    or, even easier, recall your half-angle formula to show that this is

    1/2 (1-cos(32x))

  • Calculus -

    If i use integration by parts can i take out 1/4 from the integral and make u=sin^2(16x) and dv=dx ?

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