Originally Posted by

**Yip**
This integral is just an exercise in substitution lol

I'll use the notation I[a,b]...dx to represent the definite integral from a to b wrtx.

Let J=I[0,pi/2]log(sinx)dx

Let x=pi/2-t in the integral

dx=-dt

I[0,pi/2]log(sinx)dx=I[pi/2,0]-log(cost)dt

=I[0,pi/2]log(cost)dt=I[0,pi/2]log(cosx)dx

(this can be seen from letting t=x. For the rest of this proof, I wont use the intermediate sub, but just write it out in terms of x, as the t sub is just a dummy variable)

Since I[0,pi/2]log(sinx)dx=I[0,pi/2]log(cosx)dx,

2J=I[0,pi/2]log(sinxcosx)dx

=I[0,pi/2]log(0.5sin2x)dx

=I[0,pi/2]log(0.5)dx+I[0,pi/2]log(sin2x)dx

=(-pi/2)log2+I[0,pi/2]log(sin2x)dx

=(-pi/2)log2+0.5I[0,pi]log(sinx)dx

=(-pi/2)log2+0.5(I[0,pi/2]log(sinx)dx+I[pi/2,pi]log(sinx)dx)

=(-pi/2)log2+0.5(J+I[0,pi/2]log(sinx)dx) (note:I[pi/2,pi]log(sinx)dx=I[0,pi/2]log(sinx)dx through the substitution x=pi-t)

=(-pi/2)log2+J

J=I[0,pi/2]log(sinx)dx=(-pi/2)log2

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