github twitter email rss
Maths Notes: Integration by Parts
Mar 3, 2017
2 minutes read

This is part of my notes for the Edexcel C4 exam, which I will be taking in 2017

Derivation

Given the product rule for differentiation is

$$ \frac{d}{dx}\left(uv\right) = v\frac{du}{dx} + u\frac{dv}{dx} $$

we can use this to derive a formula for a product rule while integrating.

First, rearrange in terms of \(u\frac{dv}{dx}\)

$$ u\frac{dv}{dx} = \frac{d}{dx}\left(uv\right) - v\frac{du}{dx} $$

Then, integrate each term in terms of x

$$ \int u\frac{dv}{dx} = \int \frac{d}{dx}\left(uv\right) - \int v\frac{du}{dx} $$

This can now be simplified to

$$ \int u\frac{dv}{dx} = uv - \int v\frac{du}{dx} $$

This is the integration by parts formula.

Use

The formula can be used to to swap any integral of the form \(\int u\frac{dv}{dx}dx\) with something easier to integrate -> \(\int v\frac{du}{dx}dx\)

A couple of examples would be prudent, so let’s integrate \(\int x\sin{x}dx\)

$$ I := \int x\sin{x}dx $$

We then need to decide on values for \(u\) and \(\frac{dv}{dx}\), as well calculate \(\frac{du}{dx}\) and \(v\) respectively. Luckily, there is a good rule for deciding which to choose. It can be remembered with the mnemonic I LATE.

Letter Meaning
I Inverse trigonometric function
L Logarithmic function
A Algebraic function
T Trigonometric function
E Exponential function

Using this, the closer a letter is to the top, the more likely it should be used as u, while the closer to the bottom the more likely it should be used for \(\frac{dv}{dx}\)

So in the case of our example, this should be quite straightforward

$$ u = x \Rightarrow \frac{du}{dx} = 1 $$

$$ v = -\cos{x} \Leftarrow \frac{dv}{dx} = \sin{x} $$

The last step is to use the formula, like so:

$$ I = -x\cos{x} - \int -\cos{x} \cdot 1 dx \\ = x\sin{x} - x\cos{x} $$

Use with definite integrals

Using this with a definite integral is similar, though limits need to be applied to both terms. For instance:

$$ I := \int_1^2 x\sin{x}dx $$

Then use the formula like usual, however limits must be applied to the uv term and to the integral.

$$ I = [-x\cos{x}]_1^2 - \int_1^2 -\cos{x} \cdot 1 dx \\ $$

This can then be evaluated like any other integral.

Important note

You may need to apply parts several times before you get an answer, so don’t be caught out if your answer requires you to use parts again.


Back to posts