SOLUTION: Pls help Use integration by parts to find ∫ e^-x cosx dx

Let I = 
 u = cos(x) dv = e-xdx
du = -sin(x) v = -e-x

I = uv - ∫vdu = cos(x)[-e-x] - ∫[-e-x][-sin(x)]
I = -e-xcos(x) - ∫e-xsin(x)dx

Let J = , then 

I = -e-xcos(x) - J

Now we find J:

J = 

u = sin(x)    dv = e-xdx
du = cos(x)    v = -e-x

J = uv - ∫vdu = sin(x)[-e-x] - ∫[-e-x][cos(x)]
J = -e-xsin(x) - ∫e-xcos(x)
J = uv - ∫vdu = sin(x)[-e-x] - ∫[-e-x][cos(x)]
J = -e-xsin(x) + ∫e-xcos(x)
J = -e-xsin(x) + I

Substitute that for J in

 I = -e-xcos(x) - J
 I = -e-xcos(x) -(-e-xsin(x) + I)
 I = -e-xcos(x) + e-xsin(x) - I
2I = -e-xcos(x) + e-xsin(x)
2I = e-x[-cos(x) + sin(x)]
2I = e-x[sin(x) - cos(x)]

Final answer: 



Edwin