What's the integral of e^x*cos(2x)?

The integral of e^x*cos(2x) is (1/5)*e^x*(cos(2x)+0.5*sin(2x))+C.

To solve this integral, we can use integration by parts. Let u=e^x and dv=cos(2x)dx. Then du/dx=e^x and v=(1/2)*sin(2x). Using the formula for integration by parts, we have:

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

Next, we can use integration by parts again, with u=e^x and dv=sin(2x)dx. Then du/dx=e^x and v=-(1/2)*cos(2x). Using the formula for integration by parts, we have:

∫e^x*cos(2x)dx = e^x*(1/2)*sin(2x) - e^x*(1/4)*cos(2x) + ∫(1/4)*cos(2x)*e^x dx

We can repeat this process one more time, with u=e^x and dv=cos(2x)dx. Then du/dx=e^x and v=(1/2)*sin(2x). Using the formula for integration by parts, we have:

∫e^x*cos(2x)dx = e^x*(1/2)*sin(2x) - e^x*(1/4)*cos(2x) + e^x*(1/8)*sin(2x) - ∫(1/8)*sin(2x)*e^x dx

Simplifying this expression, we get:

∫e^x*cos(2x)dx = (1/2)*e^x*sin(2x) - (1/4)*e^x*cos(2x) + (1/8)*∫e^x*sin(2x)dx

Solving for the integral of e^x*sin(2x), we can use integration by parts again, with u=e^x and dv=sin(2x)dx. Then du/dx=e^x and v=-(1/2)*cos(2x). Using the formula