Continuing on from what you have, that integral becomes
My bank of integral tricks isn't really that large, but if I try adding those two integrals up I don't get anywhere either..
And wolfram does give an answer. pi log2.
Someone solved using elementary methods on one of the previous pages.Continuing on from what you have, that integral becomes
And this integral (ignoring the -2 at the front) evaluates to , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.
I've seen people do it by using int f(pi/2 - x) = int f(x)Continuing on from what you have, that integral becomes
And this integral (ignoring the -2 at the front) evaluates to , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.
You are mostly right, not the last line though. Will fix up my eqn which isn't displaying for some reason.I've seen people do it by using int f(pi/2 - x) = int f(x)
giving I = int 0 to pi/2 of ln(costheta) dtheta
Then adding, giving 2I = int 0 to pi/2 of ln(sin2theta) - ln2 dtheta
Then using u = 2theta, 2I = int 0 to pi of ln(sinu) - ln2 du
And then they evaluate the int 0 to pi of ln(sinu) to be zero because it's an odd around theta = pi/2 or something
Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.Someone solved using elementary methods on one of the previous pages.
Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?
You use dummy variabales by letting u=2xWhy is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?
Sean did it just up there^Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.
Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?
I remember that proof now. A bit ashamed that it didn't stick with me first time I read it. It's neat proof.Sean did it just up there^
Ah that's cool
You should of course be painstakingly slow in the jumps in your working for your HSC exams.
No worries. Remember the pattern, because of log laws you can often exploit the x -> a-x substitution in ways like this.Ah that's cool
Thanks
Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?
You should of course be painstakingly slow in the jumps in your working for your HSC exams, I am just lazy / students trying to fill in blanks is actually decent exercise.
How wouldyou prove that it's the same besides using substitution: u = pi/2 - x?Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?
Bruh its a sin graph we know that its the same area...How wouldyou prove that it's the same besides using substitution: u = pi/2 - x?
lol that's a very articulate proofBruh its a sin graph we know that its the same area...
But why is the area under sin(x) between 0 and pi/2 the same as that between pi/2 and pi?Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?