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#