Question

Having a bit of trouble with \[\int_0^{\pi/2}\ln(\sin(x))dx\]. I'll post the solution given and the part that confuses me. Any help would be much appreciated!

Answer 1

Here's the given solution.

Answer 2

What I don't get is why \[\int_0^{\pi/2}\ln(\sin(x))dx = \int_0^{\pi/2}\ln(\cos(x))dx\]when it's clearly stated that this was achieved using \(x = \frac{\pi}{2} - u\). \[\int_0^{\pi/2}\ln(\cos(x))dx = \int_0^{\pi/2}\ln(\cos(u))du\]just doesn't seem to work, as I've seen on other proofs of this on other sites.

Answer 3

NOTE: That's what it meant by recommended substitution on the actual problem.

Answer 4

oooh i didn't get it at all!

Answer 5

now i see
very clever

Answer 6

put \(x=\frac{\pi}{2}-u\) \(du=-dx\) and also \(u(0)=\frac{\pi}{2}, u(\frac{\pi}{2})=0\)
you get that the integrals are the same

Answer 7

x=pi/2-u
dx=-du
integrate(ln(sin(x))dx) from x=0 to x=pi/2
=integrate(ln(cos(u))*(-du)) from x=0 to x=pi/2
=-integrate(ln(cos(u)) from u=pi/2 to u=0
=integrate(ln(cos(u)) from u=0 to u=pi/2

Answer 8

Oh! I think that I get it now. Is because the boundaries are specifically 0 and pi/2?

Answer 9

right, otherwise it would not work

Answer 10

Ah ok. Thank you so much! Just been bothering me for like an hour now. :)

Answer 11

yeah, without that hint it bothered me too