So I have the following integration: $\int{e^{-x}\sin{x}}$
I let u be the $e^{-x}$ and $dv = \sin{x} dx$
After doing the integration by parts I get:$(e^{-x})(-\cos{x})-\int{-e^{-x}*(-\cos{x})}$
From what it looks like, if I continue integration by parts, will it just continue going on forever?
$\endgroup$ 26 Answers
$\begingroup$For integration by parts, you will need to do it twice to get the same integral that you started with. When that happens, you substitute it for L, M, or some other letter. So we start by taking your original integral and begin the process as shown below.
Now using the formula
We get the following sequence, and we do another round of integration by parts.
So we end up the integral part that is the same as your original equation like so:
Now we substitute your original integral with the letter L like so...
...and replace every occurrence so the equation looks like the following:
Then we solve for L as shown below.
So your complete solution is this:
Hope this helps.
$\endgroup$ $\begingroup$If you continue (do integration by part for $\int e^{-x}\cos x\,dx$) you will get $\int e^{-x}\sin x \,dx= f(x) - \int e^{-x}\sin x\, dx$, therefore $\int e^{-x}\sin x\, dx= \frac12 f(x)$.
Every time you integrate by part you will get an extra minus, but you integrating $\sin x$ twice get one minus, that's why in this case doing integration twice works.
Edit: You also made a mistake. (Editted) $\int udv = uv -\int vdu$, so you should get
$$\int e^{-x} \sin x \,dx = -e^{-x}\cos x - \int e^{-x} \cos x \, dx$$
$\endgroup$ 2 $\begingroup$$$
\begin{align}
&\int e^{-x}\sin(x)\,\mathrm{d}x\tag{1}\\
&=-\int\sin(x)\,\mathrm{d}e^{-x}\tag{2}\\
&=-e^{-x}\sin(x)+\int e^{-x}\cos(x)\,\mathrm{d}x\tag{3}\\
&=-e^{-x}\sin(x)-\int\cos(x)\,\mathrm{d}e^{-x}\tag{4}\\
&=-e^{-x}\sin(x)-e^{-x}\cos(x)-\int e^{-x}\sin(x)\,\mathrm{d}x\tag{5}\\
&=-\frac12e^{-x}(\sin(x)+\cos(x))\tag{6}
\end{align}
$$
Explanation:
$(2)$: $-\mathrm{d}e^{-x}=e^{-x}\,\mathrm{d}x$
$(3)$: integrate by parts
$(4)$: $-\mathrm{d}e^{-x}=e^{-x}\,\mathrm{d}x$
$(5)$: integrate by parts
$(6)$: average $(1)$ and $(5)$
You actually regenerate the original integral: $$ I = \int e^{-x} \sin{x} \, dx = -e^{-x} \sin{x} + \int e^{-x} \cos{x} \, dx \\ = -e^{-x} \sin{x} -e^{-x} \cos{x} -(-1)\int e^{-x} (-\sin{x}) \, dx \\ = -e^{-x} \sin{x} -e^{-x} \cos{x} - I, $$ so $$ I = \frac{1}{2}(-e^{-x} \sin{x} -e^{-x} \cos{x}) + C. $$
Or, you can go the other way: $$ I = \int e^{-x} \sin{x} \, dx = e^{-x}(-\cos{x} ) - \int (-e^{-x})(-\cos{x}) \, dx \\ = -e^{-x} \cos{x} -\int e^{-x} \cos{x} \, dx \\ = -e^{-x} \cos{x} - e^{-x}\sin{x} -(-1) \int (-e^{-x}) \sin{x} \, dx \\ = -e^{-x} \cos{x} - e^{-x} \sin{x} - I $$
$\endgroup$ $\begingroup$$$\int e^{-x}\sin x\,dx= \frac{1}{2i}\int e^{-x}(e^{ix}-e^{-ix})\,dx= \frac{1}{2i}\left(\frac{e^{(-1+i)x}}{-1+i}-\frac{e^{(-1-i)x}}{-1-i}\right)+C= \frac{e^{-x}}{4i}\left(e^{ix}(-1-i)-e^{-ix}(-1+i)\right)+C= \frac{e^{-x}}{2}\left(-\frac{e^{ix}-e^{-ix}}{2i}-\frac{e^{ix}+e^{-ix}}{2}\right)+C=\frac{e^{-x}}{2}\left(-\sin x -\cos x\right)+C$$
$\endgroup$ 2 $\begingroup$perhaps a less painful way takes advantage of the fact that we can do the integration by parts along two routes. first clear the negative by noting that if
$$ f(x) = \int^x e^{-t}\sin t dt $$ then $$ f(-x) = \int^{-x} e^{-t}\sin t dt = \int^x e^t\sin (-t) d(-t) =\int^x e^t \sin t dt $$ now, by the first route $$ f(-x) = [e^t \sin t]^x - \int^x e^t \cos t dt \tag{1} $$ and by the second route $$ f(-x) = [-e^t \cos t]^x + \int^x e^t \cos t dt \tag{2} $$ adding (1) and (2) $$ 2f(-x) = e^x(\sin x - \cos x) +C $$ so $$ f(x) = \frac12 e^{-x}( \sin(-x) -\cos(-x)) +C \\ = -\frac12 e^{-x}(\sin x + \cos x) + C $$
$\endgroup$
