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

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Answer 1 · 2018-06-18T21:42:53

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

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

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

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

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

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

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

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

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

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