Integration by Parts – integrating products

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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:

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

When we integrate a product we choose one of the functions to be $u$ and the other function to be $\frac{dv}{dx}$. We then find $\frac{du}{dx}$ and $v$ 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 $u$ and $\frac{dv}{dx}$ so as to make the integral on the right a more simple one to find. For example, consider $\int x\cos(x)dx$. We choose $u=x$ and $\frac{dv}{dx}=\cos(x)$ as this will simplify the integral on the right hand side in the formula. It follows that $\frac{du}{dx}=1$ and $v=\sin(x)$ (we leave the integration constant until the end) and so:

$\int x\cos(x)\,dx=x\sin(x)-\int \sin(x)\times 1 \,dx =x\sin(x)+\cos(x)+c$

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

$\int_a^b \left(u\frac{dv}{dx}\right)dx=\left[uv\right]_a^b-\int_a^b v\frac{du}{dx}dx$

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 $u$ and $\frac{dv}{dx}$ initially.

Derivation of the Formula

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

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

where $u$ and $v$ are both functions of $x$. It follows that

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

We can rearrange this to get:

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

The trick for integrating $\ln(x)$

Although we know that $\ln(x)$ differentiates to $\frac{1}{x}$, we don’t yet know how to integrate $\ln(x)$. There is a trick to this one – memorise it! Treat. $\ln(x)$ as a product of one and itself. That is, write it as $\ln(x)=\ln(x)\times 1$. We recognise this as a product and we can use integration by parts to integrate it. Let $u=\ln(x)$ and $\frac{dv}{dx}=1$. It follows that $\frac{du}{dx}=\frac{1}{x}$ and $v=x$. Hence,

$\begin{array}{l}\int \ln(x)dx&=&\int \ln(x)\times 1\,dx\hspace{5pt}=\hspace{5pt}x\ln(x)-\int x\times\frac{1}{x}dx\hspace{5pt}=\hspace{5pt}x\ln(x)-\int 1\,dx\\&=&x\ln(x)-x+c\end{array}$

Choosing $u$ and $\frac{dv}{dx}$ around the other way round would not work as choosing $\frac{dv}{dx}=\ln(x)$ would require integrating $\ln(x)$ – this is what we are trying to do anyway. We can check this by differentiating using the product rule where necessary:

$\frac{d}{dx}\left(x\ln(x)-x+c\right)=1\times\ln(x)+x\times\frac{1}{x}+1=\ln(x)$.

When integrating other products that include $\ln(x)$ it is usually best to choose $u=\ln(x)$ as choosing it as $\frac{dv}{dx}$ will require integrating $\ln(x)$ and will complicate the integral. See Example 1.