What's the integral of (e^x)*cos(x)?

The integral of (e^x)*cos(x) is (1/2)*(e^x)*(sin(x) + cos(x)) + C.

To solve this integral, we can use integration by parts. Let u = e^x and dv = cos(x) dx. Then du/dx = e^x and v = sin(x). Using the integration by parts formula, we have:

∫(e^x)*cos(x) dx = e^x*sin(x) - ∫e^x*sin(x) dx

Now we can use integration by parts again, with u = e^x and dv = sin(x) dx. Then du/dx = e^x and v = -cos(x). Plugging this into the formula, we get:

∫(e^x)*cos(x) dx = e^x*sin(x) - e^x*cos(x) + ∫e^x*cos(x) dx

We can see that we have a similar integral to what we started with, but with a minus sign in front of it. So we can substitute this back into the equation and simplify:

∫(e^x)*cos(x) dx = e^x*sin(x) - e^x*cos(x) + ∫e^x*cos(x) dx
∫(e^x)*cos(x) dx = e^x*sin(x) - e^x*cos(x) + ∫(e^x)*cos(x) dx - ∫(e^x)*cos(x) dx
2∫(e^x)*cos(x) dx = e^x*sin(x) - e^x*cos(x)
∫(e^x)*cos(x) dx = (1/2)*(e^x)*(sin(x) + cos(x)) + C

Therefore, the integral of (e^x)*cos(x) is (1/2)*(e^x)*(sin(x) + cos(x)) + C.