Question

How do you evaluate `int e^(6x)sin(e^(2x))dx` using substitution and and tabular integration?

Answer 1

`inte^(6x)sin(e^(2x))dx=e^(2x)sin(e^(2x))+cos(e^(2x))-e^(4x)/2cos(e^(2x))+C`
`C in RR`

Explanation:

`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!