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)]#
