How Do I Solve ∫e^-x cos(2x)dx Using Integration by Parts?

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SUMMARY

The integral ∫e^-x cos(2x)dx can be solved using integration by parts, where u = e^-x and dv = cos(2x)dx. This leads to the equation ∫udv = (e^-x)(1/2 sin(2x)) + 1/2∫sin(2x)e^-x dx. By applying integration by parts again on the integral ∫sin(2x)e^-x dx, one can derive an equation that includes the original integral on both sides, allowing for algebraic manipulation to isolate and solve for ∫e^-x cos(2x)dx.

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Homework Statement



∫e^-x cos(2x)dx

Homework Equations


I'm trying integration by parts and I set u=e^-x and dv=cos(2x)

The Attempt at a Solution


I got to where ∫udv= (e^-x)(1/2 sin(2x))+1/2∫sin(2x)e^-x
I am trying to run through a second time and I'm a little stuck
 
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Take ∫sin(2x)e-x dx separately.

u= e-x du=?

dv= sin(2x)dx v =?
 
Continuing as you are, you should get an equation with ∫e^-x cos(2x)dx on one side and the same on the other side, plus some other terms. Move both integral terms to one side of the equation and solve algebraically for the integral.
 

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