Integration by parts of $\cos(x)e^{-x}dx$ I do the integral but I end up getting the original $\cos(x)e^{-x}dx$ on both sides and canceling them out resulting in no solution.
Can I get a step by step break down of how to solve?

Using $e^{-x}=dv$ and $cos(x)=u$: $$-e^{-x}cos(x)-\int e^{-x}sin(x)$$ $$\int e^{-x}sin(x)=-e^{-x}sin(x)-\int -e^{-x}cos(x)$$ Be careful with the signs, and you will get an expression like this: $$2\int e^{-x}cos(x)=-e^{-x}cos(x)+e^{-x}sin(x)$$
Divide by two, and that is your solution.
