#inte^(6x)sin(e^(2x))dx#
Let #t=e^(2x)#
#dt=2e^(2x)dx#
#inte^(6x)sin(e^(2x))dx=int1/2(2e^(2x)e^(4x)sin(e^(2x))) dx#
#=1/2intt²sin(t)dt#
Using Integration by parts :
#intg'(t)f(t)dt=[f(t)g(t)]-intf'(t)g(t)dt#
There :
#f(t)=t²# #g'(t)=sin(t)#
#f'(t) =2t# #g(t)=-cos(t)#
So:#1/2intt²sin(t)dt=1/2([-t²cos(t)]+2inttcos(t)dt)#
Using again integration by parts :
#f(t)=t# #g'(t)=cos(t)#
#f'(t)=1# #g(t)=sin(t)#
So:#1/2([-t²cos(t)]+2inttcos(t)dt)=1/2([-t²cos(t)]+2([tsin(t)]-intsin(t)dt))#
#=1/2([-t²cos(t)]+[2tsin(t)]+[2cos(t)])#
#=tsin(t)+cos(t)-t²/2cos(t)+C#
#C in RR#
Finally, because #t=e^(2x)#:
#=e^(2x)sin(e^(2x))+cos(e^(2x))-e^(4x)/2cos(e^(2x))+C#
#C in RR#
\0/ Here's our answer!
