Will Huxtable

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Maths Notes: Integration by Parts
Mar 3, 2017

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.

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