Integration by parts of $\cos(x)e^{-x}dx$

Question

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?

Answer 1

First, set $u = \cos x$ and $dv = e^{-x} dx$, so $du = - \sin x \,dx$ and $v = - e^{-x}$. We get $$\int \cos (x) e^{-x} \,dx = (\cos (x))(-e^{-x}) - \int (-\sin (x))(-e^{-x})\, dx$$. Now, set $u = - \sin x$ and $dv = -e^{-x}\,dx$ to get $du = - \cos x\, dx $ and $v = e^{-x}$. This gives us $$\int \cos (x) e^{-x} \,dx = (\cos(x))(-e^{-x}) - (-\sin(x))(e^{-x}) - \int \cos(x)e^{-x}\, dx.$$ Hence, $$ \int \cos(x) e^{-x}\, dx = \frac{e^{-x}}{2}(\sin(x) - \cos(x)) + C.$$

Answer 2

$$\int\cos xe^{-x}dx=Re\int e^{ix}e^{-x}dx=Re\int e^{x(i-1)}dx$$

Answer 3

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.

Answer 4

Letting $u = e^{-x}$ and $dv = \cos x \, dx$ so that $du = -e^{-x} \, dx$ and $v = \sin x$, we obtain: \begin{align*} \int e^{-x}\cos x \, dx &= (e^{-x})(\sin x) - \int (\sin x)(-e^{-x} \, dx) \\ &= e^{-x}\sin x + \int e^{-x}\sin x \, dx \end{align*} Letting $u = e^{-x}$ and $dv = \sin x \, dx$ so that $du = -e^{-x} \, dx$ and $v = -\cos x$, we obtain: \begin{align*} \int e^{-x}\cos x \, dx &= e^{-x}\sin x + (e^{-x})(-\cos x) - \int(-\cos x)(-e^{-x} \, dx)\\ &= e^{-x}(\sin x -\cos x) - \int e^{-x}\cos x \, dx \end{align*} which contains the original integral! Collecting like terms and dividing by $2$, we conclude that: $$ \int e^{-x}\cos x \, dx = \frac{1}{2}e^{-x}(\sin x -\cos x) + C $$