Question

How do you find the Integral of `e^(3x)* cos( 3 x ) dx`?

Answer 1

`I=(1/3e^(3x)sin3x+1/3e^(3x)cos3x)/2`

Explanation:

`I=inte^(3x)cos(3x)dx`

Apply Integration by Parts

`color(green)(intudv=uv-intvdu)" "rarr "` `color(blue)("the trigonometric part will be " dv " in this case" color(red)( " i.e " cos3xdx=dv)`

`I=1/3e^(3x)sin3x-intcancel(1/3)sin3x*(cancel(3)e^(3x))dx`

Apply integration by Parts once more

`I=1/3e^3xsin3x+1/3e^(3x)cos3x-cancel(1/3)intcancel(3)e^(3x)cos3xdx`

`I=1/3e^(3x)sin3x+1/3e^(3x)cos3x-color(blue)(I`

`color(blue)("since " I=inte^(3x)cos3xdx`

`2I=1/3e^(3x)sin3x+1/3e^(3x)cos3x`

`I=(1/3e^(3x)sin3x+1/3e^(3x)cos3x)/2`