What is #int_0^pi sin^2(lnx)#?

1 Answer
Jan 9, 2016

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)]#