How do you find the integral of $e^(2x) cos4x dx$ ?

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Answer 1 · 2015-10-10T02:32:37

Integrate by parts twice using $u=e^(2x)$ both times.

$I = int e^(2x) cos4x dx$

$u= e^(2x)$ and $dv = cos4x dx$

$I = uv-intvdu$

$= 1/4e^(2x)sin4x-1/2inte^(2x)sin4xdx$

Now take $u = e^(2x)$ and $dv = sin4xdx$

$I=1/4e^(2x)sin4x-1/2[-1/4e^(2x)cos4x +1/2inte^(2x)cos4xdx]$

$I = 1/4e^(2x)sin4x+ 1/8 e^(2x)cos4x - 1/4 I$

Solve for $I = 1/5e^(2x)sin4x+ 1/10 e^(2x)cos4x +C$

How do you find the integral of e^(2x) cos4x dxe^(2x) cos4x dx?