We have:
#inttsin(mt)dt#, where #m# is presumably real, and does not equal zero. We say that #m!=0# and #m inRR#.
According to integration by parts, #intf(x)g(x)dx=f(x)intg(x)dx-intf'(x)(intg(x)dx)dx#
Here, #f(t)=t# and #g(t)=sin(mt)#. We have:
#tintsin(mt)dt-int(intsin(mt)dt)dt#
We need to compute #intsin(mt)dt#. Using integration by substitution,
#intf(g(x))g'(x)dx=intf(u)du#, where #u=g(x)#. Here, #u=mt#, and #u'=m#, so we can say:
#1/m intsin(mt)mdt#
#1/m intsin(u)du#
#1/m*-cos(u)#
#-cos(mt)/m#. Inputting:
#-(tcos(mt))/m-int-cos(mt)/mdt#
#-(tcos(mt))/m+1/m intcos(mt)dt#
Using integration by substitution, where #u=mt# again, we have:
#-(tcos(mt))/m+1/m(1/m intcos(mt)mdt)#
#-(tcos(mt))/m+1/m(1/m intcos(u)du)#
#-(tcos(mt))/m+1/m(1/m sin(mt))#
#sin(mt)/m^2-(tcos(mt))/m#
#(sin(mt)-mtcos(mt))/m^2+C#
