How do you find the Integral of e^(3x)* cos( 3 x ) dxe3xcos(3x)dx?

1 Answer
Jun 18, 2018

I=(1/3e^(3x)sin3x+1/3e^(3x)cos3x)/2I=13e3xsin3x+13e3xcos3x2

Explanation:

I=inte^(3x)cos(3x)dxI=e3xcos(3x)dx

Apply Integration by Parts

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