Calculus Tutorial

Intro

Integration by parts is an important and powerful integration technique that captures the idea of using the product rule of differentiation backwards.

Let’s recall the product rule of differentiation:

$\frac{d}{dx}(f(x)*g(x)) = g(x)*\frac{d}{dx}(f(x)) + f(x)*\frac{d}{dx}(g(x))$

Now suppose we were trying to integrate an expression that could be written as g(x)*f'(x). We tried power rule, u-substitution, and scored our table of integrals, but to no avail. Then in a flash of insight, we recall the product rule of our differentiation days long gone by! Since the integral and antiderivative are one and the same, the product rule tells us that the integral of g(x)*f'(x) + f'(x)*g(x) is f(x)*g(x).

$\int{(g(x)*\frac{d}{dx}(f(x)) + f(x)*\frac{d}{dx}(g(x)))dx} = f(x)*g(x)$

Separating the integral on the left:

$\int{g(x)*\frac{d}{dx}(f(x))dx + \int{f(x)*\frac{d}{dx}(g(x))dx}} = f(x)*g(x)$

Now isolate the integral of g(x)*f'(x)…

$\int{g(x)*\frac{d}{dx}(f(x))dx} = f(x)*g(x) - \int{f(x)*\frac{d}{dx}(g(x))dx}$

And viola, integration by parts. A more conventional way to write this is in terms of of u and v (A rather strange choice given the similarity of their appearance, but apparently the grand patriarchs of calculus had rather impeccable handwriting.) rather than f(x) and g(x), and to cancel the dos in the numerator and denominator giving:
$\int{udv} = uv - \int{vdu}$
I personally have a hard time remembering where u and v go because they look so similar, so I usually just go through the derivation in my head when I approach a problem, but the uv notation is what you will see pretty much everywhere.

Now, while any integral can be expressed in this form, it is not necessarily any easier to solve. The key for the integration by parts to be helpful is for the integral of udv to work out nicely. The key to making sure that this happens as often as possible is to choose our us and vs wisely.

There is a handy acronym for figuring out which of the two parts of the product we are integrating should be u, the one that stays the same in the product uv.

The acronym is LIPET, like lipid from biology.
L – Logarithm
I – Inverse Trig
P – Polynomial (including 1 and x)
E – Exponential (like e^x)
T – Trig

LIPET establishes an order of precedence for choosing u. The function in your integrand that is higher up on the list is u, and the function that is lower on the list is dv. For example, if we were trying to integrate e^x*arctan(e^x)dx, then u would be arctan(e^x) and du would be e^xdx.

This rule is not hard and fast, and is simply a trick that is usually helpful. There are other acronyms that make minor changes to the position of a few of the functions, but here is one thing that is pretty true: having 1/x (derivative of ln(x)) in an integral is much nicer than having cos(x) (derivative of sin(x)) in an integral. This is the basic scheme behind LIPET and its cousins.

Sample Problem

Here is a classic integration by parts problem:
$\int{e^x\sin{x}dx}$

Solution

By LIPETS, e^x should stay the same (u), and sin(x)dx should be integrated (dv).

So we have:
$\int{e^x\sin{x}dx} = e^x\int \sin{x}dx - \int((\int \sin{x}dx)\frac{d}{dx}(e^x)dx)$
$\int{e^x\sin{x}dx} = -e^x\cos{x} + \int{e^x\cos{x}}}dx}$

Applying integration by parts a second time gives:
$\int{e^x\sin{x}dx} = -e^x\cos{x} + e^x\sin{x} - \int{e^x\sin{x}dx}$

Now we can add the integral of e^xsin(x)dx to both sides:
$2\int{e^x\sin{x}dx} = -e^x\cos{x} + e^x\sin{x}$

Divide by 2 and……
$\int{e^x\sin{x}dx} = \frac{1}{2}(-e^x\cos{x} + e^x\sin{x}) = \frac{1}{2}e^x(\sin{x} - \cos{x}) + C$

That’s all folks and I’ll see you next time.