*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.