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!
