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Calculus

posted by on .

I need help with this integral.

w= the integral from 0 to 5

24e^-6t cos(2t) dt.

i found the the integration in the integral table.

(e^ax/a^2 + b^2) (a cos bx + b sin bx)

im having trouble finishing the problem from here.

  • Calculus - ,

    Write cos(x) as the real part of
    exp(ix)

    The integral of exp(ax)cos(bx) is then the real part of the integral of
    exp[(a + ib)x]

    The integral is:

    1/(a+ib) exp[(a+ib)x] =

    (a-ib)/(a^2 + b^2) exp(ax)*
    [cos(bx) + i sin(bx)]

    The real part of this is:

    exp(ax)/(a^2 + b^2)
    [a cos(bx) + b sin(bx)]

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