Calculus

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Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help.

  • Calculus -

    1/2 ∫ 1/(sin(x/2)cos(x/2))dx
    let
    u = sin(x/2)
    du = 1/2 cos(x/2) dx
    or, dx = 2/cos(x/2) du

    Then you have

    1/4 ∫1/u 2/(cos(x/2))dx
    = 1/4 ∫ 1/u du
    = 1/4 ln(sin(x/2)) + C

    Now, we all know that
    ∫ csc(x)dx = -ln(cscx + cotx)

    so what gives here?

    1/4 ln(sin(x/2))
    = -1/4 ln(1/sin(x/2))
    = -1/2 ln(1/sqrt(1-cosx))
    gotta run, but I think if you manipulate things a bit and adjust the C it will work out to be the same.

    I'll check in later to make sure.

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