1. ## Integration

Integration:

I=integral:[lnsinx] boundaries a=pi/2 b=0

nudge nudge, help?

2. ## Re: Integration

This is a bitch.

I tried to do it using the parts method where i made
u = lnsinx
du = cosx/sinx

and dv = dx
v = x

Then i got I = xlnsinx - integral (xcosx/sinx)

Then i made integral(xcosx/sinx) = I(2)

u = x
du = dx

dv = sinx/cosx
v=lnsinx

I(2) = xlnxsinx - integral (lnsinx)
I(2) = xlnsinx - I

I = xlnsinx - (xlnsinx - I)
I = xlnsinx - xlnsinx + I
I = I

lol i did that three times.. sorry i wasn't much help

3. ## Re: Integration

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

4. ## Re: Integration

How did you do this:

I[0,pi/2]log(0.5sin2x)dx=I[0,pi/2]log(0.5)dx+I[0,pi/2]sin2xdx

Also, how did you do:

(-pi/2)log2+I[0,pi/2]sin2xdx=(-pi/2)log2+0.5I[0,pi]sinxdx

Probably some extremely clever trick that I don't know about

I can see genius in that working out. Please explain
I think the first example I posted is wrong though.

5. ## Re: Integration

Yeah, it was a typo on my part lol. I've edited the working accordingly.

In I[0,pi/2]log(sin2x)dx, u let x=t/2
dx=0.5dt
I[0,pi/2]log(sin2x)dx=0.5I[0,pi]log(sint)dt=0.5I[0,pi]log(sinx)dx

6. ## Re: Integration

Originally Posted by 3unitz

Where there's a will...

7. ## Re: Integration

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

Fucking genius.

Yip, how did you know to sub pi-t ? Did you just experiment, or is there a definitive method to it?

1. ###### Math Is Fun Forum / Integral challenge for masters
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