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?
 A: 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,
$$ \int \cos(x) e^{-x}\, dx = \frac{e^{-x}}{2}(\sin(x) - \cos(x)) + C.$$
A: $$\int\cos xe^{-x}dx=Re\int e^{ix}e^{-x}dx=Re\int e^{x(i-1)}dx$$
A: 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.
A: 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:
$$
\int e^{-x}\cos x  \, dx = \frac{1}{2}e^{-x}(\sin x -\cos x) + C
$$
