#f(x) = int1/x^2*sin(2ln(x))dx#
#= 1/2int2/x^2*sin(2ln(x))dx#
#u = 2ln(x)# note #u = ln(x^2)#
#du = 2/x dx#
#= 1/2int2/x*1/x*sin(2ln(x))dx#
#=1/2int1/x*sin(u) du#
#e^u = x^2#
#e^(1/2u)=x#
#e^(-1/2u) = 1/x#
#f(x)=int1/2e^(-1/2u)*sin(u) du#
By part :
#g(x) =1/2e^(-1/2u)#
#g'(x)=-1/4e^(-1/2u)#
#h'(x)=sin(u)#
#h(x)=-cos(u)#
#int1/2e^(-1/2u)*sin(u) du=-[1/2e^(-1/2u)cos(u)]-int1/4e^(-1/2u)cos(u)du#
by part again :
#g(x) = 1/4e^(-1/2u)#
#g'(x) = -1/8e^(-1/2u)#
#h'(x)=cos(u)#
#h(x)=sin(u)#
#=int1/2e^(-1/2u)*sin(u)du=-[1/2e^(-1/2u)cos(u)]-([1/4e^(-1/2u)sin(u)]+1/4int1/2e^(-1/2u)sin(u)du)#
#=int1/2e^(-1/2u)*sin(u) du=-[1/2e^(-1/2u)cos(u)]-[1/4e^(-1/2u)sin(u)]-1/4int1/2e^(-1/2u)sin(u)du#
#=4int1/2e^(-1/2u)*sin(u) du=-4[1/2e^(-1/2u)cos(u)]-4[1/4e^(-1/2u)sin(u)]-int1/2e^(-1/2u)sin(u)du#
#=4int1/2e^(-1/2u)*sin(u) du=-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)]-int1/2e^(-1/2u)sin(u)du#
#=5int1/2e^(-1/2u)*sin(u) du=-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)]#
#=int1/2e^(-1/2u)*sin(u) du=1/5(-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)])#
Substitute back : #e^(-1/2u) = 1/x#
So we have :
#=(-1/(5x)(2cos(u)+sin(u)))+C#
