In a message dated 3/27/2018 12:35:31 PM Eastern Standard Time, anlytcphil@aol.com writes:
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