How to do a u-substitution

Within the realm of integration, all techniques can be understood in terms of their differentiation counterparts. U-substitutions are no different, but the technique of which it is the inverse may not be immediately obvious. It is in fact the inverse chain rule, which we will explain below.

Proof

The chain rule can be summarised as:

$$
\frac{d}{dx} f(g(x)) = f’(g(x)) \times g’(x)
$$

The nested functions can be confusing, so think of the $f’(g(x))$ as differentiating the outer function $f(x)$ with respect to the inner function $g(x)$. As an example:

$$
\frac{d}{dx} sin(x^2)
$$

We will differentiate the sine function with respect to $x^2$, giving $cos(x^2)$, and then we will multiply by $2x$:

$$
= cos(x^2) \times 2x
$$

This is important for u-substitution. Let’s consider integrating both sides of the chain rule expression:

$$
\int \frac{d}{dx} f(g(x)) \ dx = \int f’(g(x)) \times g’(x) \ dx
$$

Left hand side is an integral of a derivative, which just cancels down to $f(g(x))$. Switching LHS and RHS:

$$
\int f’(g(x)) \times g’(x) \ dx = f(g(x))
$$

And this is our u-substitution expression. We can see that to integrate a nested function, we need the derivative of the inside function $g’(x)$ on the end.

To make things easier, we will substitute $u = g(x)$ when we solve questions like this; this is where the technique gets its name. But when we substitute $u$, we need to also substitute $du$, as we must integrate functions in terms of $u$ with respect to $u$, not $x$. This is done as follows:

$$
\begin{equation} \label{eq1}
\begin{split}
g’(x) & = \frac{d}{dx}u \\
g’(x)dx & = du
\end{split}
\end{equation}
$$

So thus our $du$ will ‘swallow up’ the derivative function $g’(x)$. Once we are in terms of $u$, we can integrate like any other function:

$$
\begin{equation} \label{eq2}
\begin{split}
\int f’(u)du & = f(u) + C
\end{split}
\end{equation}
$$

We can then finally substitute $u = g(x)$ back in to our answer, to get it in terms of $x$:

$$
f(u) + C = f(g(x)) + C
$$

Example

Lets consider:

$$
\int xcos(x^2)dx
$$

As we have nested functions, we will integrate by u-substitution. When we determine what to set $u$ to be, we must consider that we must have its derivative on the outside (remember that $g’(x)$ on the end). So let’s set it to be $x^2$:

$$
\text{Let}\ u = x^2
$$

Now we need to calculate the derivative of $u$ with respect to $x$:

$$
\begin{equation} \label{eq3}
\begin{split}
\frac{du}{dx} & = 2x \\
xdx &= \frac{1}{2}\ du
\end{split}
\end{equation}
$$

And finally we can substitute in to get a new integral in terms of $u$:

$$
\begin{equation} \label{eq4}
\begin{split}
& \int \frac{1}{2}cos(u)du \\
& = \frac{1}{2}sin(u) + C \\
& = \frac{1}{2}sin(x^2) + C
\end{split}
\end{equation}
$$

A word about bounds

This is where most students go wrong. If we have a definite integral (one which lower and upper bounds), we must write these in terms of $u$ as well. To do this, we subsitute the bounds for $x$ into our $u(x)$ function (the statement in which we let $u$ be some function of $x$) to compute our new bounds.

For example, if the previous integral was actually:

$$
\int^2_1 xcos(x^2)dx
$$

Our new bounds would be, based on the statement $u = x^2$:

$$
\begin{equation} \label{eq5}
\begin{split}
\text{Upper} = 2^2 = 4 \\
\text{Lower} = 1^2 = 1
\end{split}
\end{equation}
$$

So our integral in terms of $u$ is now:

$$
\int^4_1 \frac{1}{2}cos(u)du \\
$$