#I=int_0^(pi/2) Ln(sinx)*dx#
=#[xLn(sinx)]_0^(pi/2)-int_0^(pi/2) xcotx*dx#
=#0-int_0^(pi/2) (x*dx)/tanx#
=#-int_0^(pi/2) (arctan(tanx)*dx)/tanx#
After using Feynman's trick,
#I(b)=-int_0^(pi/2) (arctan(btanx)*dx)/tanx#
#I'(b)=-int_0^(pi/2) dx/[(btanx)^2+1]#
=#-int_0^(pi/2) ([(tanx)^2+1]*dx)/([(btanx)^2+1][(tanx)^2+1])#
After using #u=tanx# and #du=[(tanx)^2+1]*dx# transformation, this integral became,
#I'(b)=-int_0^(oo) (du)/[((bu)^2+1)*(u^2+1)]#
=#1/(b^2-1)int_0^(oo) (du)/(u^2+1)#-#b^2/(b^2-1)int_0^(oo) (du)/((bu)^2+1)#-
=#1/(b^2-1)[arctanu]_0^(oo)#-#b/(b^2-1)[arctan(u/b)]_0^(oo)#
=#pi/2*[1/(b^2-1)-b/(b^2-1)]#
=#-pi/2*(b-1)/((b-1)*(b+1))#
=#-pi/2*1/(b+1)#
After integrating both sides,
#I(b)=-pi/2Ln(b+1)+C#
For vanishing original integral, set #b=0#
#I(0)=-pi/2Ln(0+1)+C#
#C-pi+2*Ln1=0#
#C=pi/2*0=0#
Thus, #I=I(1)=-pi/2Ln2#
