# Calculus

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.

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1. 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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posted by Steve

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