# What is the general method for integrating by parts?

Apr 18, 2016

$\int u$ $\mathrm{dv} = u v - \int$ $v$ $\mathrm{du}$

#### Explanation:

The trick with doing integration by parts comes in choosing your $u$ and $\mathrm{dv}$ accordingly. If you choose poorly, the problem will usually become harder.

For example, consider $\int x \sin x$ $\mathrm{dx}$.

If you let $u = \sin x$ and $\mathrm{dv} = x$ $\mathrm{dx}$ (and accordingly $\mathrm{du} = \cos x$ $\mathrm{dx}$ and $v = \frac{1}{2} {x}^{2}$) you will end up with

$\int x \sin x$ $\mathrm{dx} = \frac{1}{2} {x}^{2} \sin x - \frac{1}{2} \int {x}^{2} \cos x$ $\mathrm{dx}$.

You can try to evaluate the integral $\int {x}^{2} \cos x$ $\mathrm{dx}$ but if you make similar $u$ and $\mathrm{dv}$ choices this problem will continue to get more complicated.

Instead, for $\int x \sin x$ $\mathrm{dx}$, let $u = x$ and $\mathrm{dv} = \sin x$ $\mathrm{dx}$ (and accordingly $\mathrm{du} = 1$ $\mathrm{dx}$ and $v = - \cos x$). Then you get

$\int x \sin x$ $\mathrm{dx} = - x \cos x - \int - \cos x \cdot \left(1\right)$ $\mathrm{dx}$

$= - x \cos x + \int \cos x$ $\mathrm{dx} = - x \cos x + \sin x + C$.

This way, you can actually do the problem.

Now you are probably thinking "Is there a way to know to to make $u$ and $\mathrm{dv}$ so I don't have to go through this process of trial and error?"

The answer is yes. Use the acronym LIPET to remember the order of choosing $u$, in order of best to choose to worst to choose.

BEST
L: logarithmic stuff
I: inverse trig stuff
P: polynomialish stuff
E: exponential stuff
T: trig stuff
WORST

Trial and error should help explain why this order is helpful.