Guy, i wanted to sleep now... but the integral is so attractive...
Let's u = ln(x)
du = 1/xdx
dx = x du
x = e^u
inte^usin^2(u)du => (a)
by the way sin^2(u) = 1/2(1-cos(2u))
1/2inte^u(1-cos(2u))du
1/2(inte^udu - inte^ucos(2u)du) => (b)
We keep our interest on inte^ucos(2u)du because the first is trivial
by part :
v = e^u
dv = e^u
dw = cos(2u)
w = 1/2sin(2u)
at this point i consider intcos(2u)du as trivial too
= 1/2[sin(2u)e^u] - 1/2inte^usin(2u)du
by part again :
v = e^u
dv = e^u
dw = sin(2u)
w = -1/2cos(2u)
=-1/2[e^ucos(2u)]+1/2inte^ucos(2u)du
Now this is interesting
inte^ucos(2u)du = 1/2[sin(2u)e^u] - 1/2(-1/2[e^ucos(2u)]+1/2inte^ucos(2u)du)
inte^ucos(2u)du = 1/2[sin(2u)e^u] +1/4[e^ucos(2u)]-1/4inte^ucos(2u)du
adding 1/4inte^ucos(2u)du both side
5/4inte^ucos(2u)du = 1/2[sin(2u)e^u] +1/4[e^ucos(2u)]
inte^ucos(2u)du = 4/10[sin(2u)e^u] +1/5[e^ucos(2u)]
Introducing in (b)
1/2(inte^udu - (4/10[sin(2u)e^u] +1/5[e^ucos(2u)]))
1/2([e^u]- (4/10[sin(2u)e^u]+1/5[e^ucos(2u)]))
1/2[e^u] -2/10[sin(2u)e^u] -1/10[e^ucos(2u)]
Substitute back for u = ln(x)
1/2[x] -2/10[sin(2ln(x))x]_0^pi -1/10[xcos(2ln(x))]_0^pi
You can factorize by -1/10x and rewrite
1/2[x] -2/10[sin(2ln(x))x]-1/10[xcos(2ln(x))]
-1/10x(2sin(2ln(x))+cos(2ln(x))-5)
now apply
[-1/10x(2sin(2ln(x))+cos(2ln(x))-5)]_0^pi
=[-1/10pi(sin(2ln(pi))+cos(2ln(pi))-5)]
