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calc bc integration of parts??

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calc bc integration of parts??
by selina z - Thursday, August 11, 2022, 8:47 PM

I am having some trouble with integration with parts, the integrals with multiplicative terms.
Specifically, I am asking about the Larson Textbook chapter 8 Review #9:

(integral) x* e^(1-x) dx
indefinate integral???

similarly, #11 in the same chapter:
(integral) e^(2x) * sin(3x) dx

Is there a general formula for these kind of problems?
ex. (integral) u v dx = ????

Thank you!
Picture of Dr. Kevin Wang
Re: calc bc integration of parts??
by Dr. Kevin Wang - Friday, August 12, 2022, 1:15 AM

In general, integration by parts is based on the product rule of derivatives: $(uv)'=u'v + uv'$.  So \[u'v = (uv)' - uv',\] then \[ \int u'v dx = uv - \int uv' dx. \] The rule of thumb is that the new integrand $uv'$ should be simpler (at least not more complicated) than the original integrand $u'v$.

For $\int x e^{1-x} dx$, if we assume $u'=x$ and $v=e^{1-x}$, then $u=\dfrac{x^2}{2}$ and $v'=-e^{1-x}$, then $uv' = \dfrac{x^2 e^{1-x}}{2}$ is more complicated than the original function, which is not the direction we want. So we can try the other way: let $u'=e^{1-x}$ and $v=x$, then $u = -e^{1-x}$ and $v'=1$, so $uv' = -e^{1-x}$, and this is indeed simpler. Thus \[ \int x e^{1-x} dx = -e^{1-x}\cdot x + \int e^{1-x}dx. \] and you can go from there.

Then integral $\int e^{2x}\sin 3x \; dx$ is similar, but a little trickier. You can give it a try and carry out a couple steps, and see what you get and whether you can find a way to make it work.