# Integration by parts

We can use integration by parts when we are required to integrate a product. The Edexcel formula booklet gives the formula for integration by parts:

When we integrate a product we choose one of the functions to be and the other function to be . We then find and and plug them into the formula above. It might seem like we’ve complicated things as there are two terms on the right, but the idea is to choose and so as to make the integral on the right a more simple one to find. For example, consider . We choose and as this will simplify the integral on the right hand side in the formula. It follows that and (we leave the integration constant until the end) and so:

If the integral is a definite one we can include the limits as follows:

See Example 1. One application of integration by parts may not be enough in some examples. Even though an integral might be simplified from one application, it may take another to complete the integration – see Example 2. In some of these cases, especially trigonometric products, it may seem that you end up with the same integral again. However, it is still possible to deduce the integral from this – see Example 3. Note that in this example, it doesn’t matter which way round we choose and initially.

## Derivation of the Formula

The formula above comes from the product rule for differentiation. Recall that the product rule for differentiation:

where and are both functions of . It follows that

We can rearrange this to get:

## The trick for integrating

Although we know that differentiates to , we don’t yet know how to integrate . There is a trick to this one – memorise it! Treat. as a product of one and itself. That is, write it as . We recognise this as a product and we can use integration by parts to integrate it. Let and . It follows that and . Hence,

Choosing and around the other way round would not work as choosing would require integrating – this is what we are trying to do anyway. We can check this by differentiating using the product rule where necessary:

.

When integrating other products that include it is usually best to choose as choosing it as will require integrating and will complicate the integral. See Example 1.

## Examples

Evaluate .

Solution:

In this example, we choose (since integrating will be more complicated) and . It follows that and . Hence,

Find .

Solution:

Since the integrand is a product and there is an term, this will require integration by parts twice. Choose and and so and . It follows that

.

The final integral here is simpler but we still require another application of integration by parts. This time choose and and so and . It follows that

.

(see below as to why we use here for the integration constant). Substituting into the first integral we have:

Note that is just an arbitrary constant if is an arbitrary constant â€“ we can simply call it .

Use integration by parts to find .

Solution:

We let , we will see why below. Here we shall choose and . Hence, and and so

Applying integration by parts again with and so that and we have

Substituting back into the above we have:

Hence,

where we have used the trigonometric identities in the formula booklet. This shows that the solution is the same as that obtained on Integrating with Trigonometric Identities page. The solution using integration by parts seen here is much longer and so using identities in the beginning is preferable.

If you prefer to use integration by parts anyway note that the choice of and doesnâ€™t matter in the first application of integration by parts. However, for the second they must be chosen the opposite way round. This method will fail if you donâ€™t.

## Extra Resources

### Questions by Topic

A2IntbyPartsExamQuestions2