Integration by substitution simplifies the process of finding integrals by making a change of variables. This is used for calculating areas and volumes in engineering, solving differential equations in physics, and optimising functions in economics. Use this resource to learn how to integrate by substitution.
An expression composed of two functions can be complicated to integrate. You can simplify this by substituting a single pronumeral (like \(u\)) to represent one of the functions. By substituting variables, we can transform the integral into a simpler form.
The substitution rule
The substitution rule for integration is like the chain rule for differentiation: it breaks a complex expression into manageable parts.
The basic rule is that if \(u=g(x)\) and \(\dfrac{du}{dx}=g'(x)\), then:
\[ \int f(g(x))g'(x)dx=\int f(u)du \]
This may also be written:
\[\int f(u)\frac{du}{dx}dx = \int f(u)du\]
Using the substitution rule
A simple test may be used to see if substitution is useful.
Look at the integrand (the function being integrated) and see if one part of it is the derivative of another part multiplied by a constant. If so, substitution may be helpful.
For example, consider \(\int2x\cos(x^{2})dx\). The derivative of \(x^{2}\) is \(2x\), and so one part is the derivative of another part and substitution can be used.
Another example is the integral \(\int x\cos(x^{2})dx\). The derivative of \(x^{2}\) is \(2x\). Since \(x\) is \(\dfrac{1}{2}\) of the derivative of \(x^{2}\), substitution is possible.
Example 1 – using the substitution rule
Calculate \(\int2x\cos(x^{2})dx\).
Here, we use the second form of the substitution rule.
\[ \int f(u)\frac{du}{dx}dx=\int f(u)du \]
We let \(u=x^{2}\) since the derivative is \(2x\), a part of the integrand. So, \(\cos(x^{2})=\cos(u)=f(u)\) and \(\dfrac{du}{dx}=2x\). Substituting these in the integral gives:
\[\begin{align*} \int2x\cos(x^{2})dx & = \int\frac{du}{dx}\cos(u)dx\\
& = \int f(u)\frac{du}{dx}dx\\
& = \int f(u)du\\
& = \int\cos(u)du\\
& = \sin(u)+c+c,\,c\in\mathbb{R}\\
& = \sin(x^{2})+c
\end{align*}\]
In some other resources, you may encounter slightly different notation. For this example, \(\dfrac{du}{dx}=2x\) may be rearranged to give \(dx=\dfrac{1}{2x}du\). Then substituting for \(x^{2}\) and \(dx\) in the integral, you would get:
\[\begin{align*} \int2x\cos(x^{2})dx & = \int2x\cos(u)\frac{1}{2x}du\\
& =\int\cos(u)du\\
& =\sin(u)+c,\,c\in\mathbb{R}\\
& =\sin(x^{2})+c
\end{align*}\]
This method gives the same result and will always work. However, you should realise that \(\dfrac{du}{dx}\) is not a fraction.
Calculate \(\int x\cos(x^{2})dx\).
Again, let \(u=x^{2}\) because \(\dfrac{du}{dx}=2x\) and so \(x=\dfrac{1}{2}\dfrac{du}{dx}\).
Like in the first example, \(\cos(x^{2})=\cos(u)=f(u)\) and substituting in the integral, we get:
\[\begin{align*} \int x\cos(x^{2})dx & = \int\frac{1}{2}\frac{du}{dx}\cos(u)dx\\
& =\frac{1}{2}\int f(u)\frac{du}{dx}dx\\
& =\frac{1}{2}\int f(u)du\\
& =\frac{1}{2}\int\cos(u)du\\
& =\frac{1}{2}\sin(u)+c,\,c\in\mathbb{R}\\
& =\frac{1}{2}\sin(x^{2})+c
\end{align*}\]
Find \(\int(8x-5)^{9}dx\).
We let \(u=8x-9\), then \(\dfrac{du}{dx}=8\), so \(1=\dfrac{1}{8}\dfrac{du}{dx}\) and \((8x-5)^{9}=u^{9}=f(u)\).
Substituting in the integral, we get
\[\begin{align*} \int(8x-5)^{9}dx & = \int\frac{1}{8}\frac{du}{dx}u^{9}dx\\
& = \int\frac{1}{8}\frac{du}{dx}f(u)dx\\
& = \frac{1}{8}\int f(u)du\\
& = \frac{1}{8}\int u^{9}du\\
& = \frac{1}{8}\frac{1}{10}u^{10}+c,\,c\in\mathbb{R}\\
& = \frac{1}{80}(8x-5)^{10}+c
\end{align*}\]
Note that \((8x-5)^{9}=1\times(8x-5)^{9}\).
Find \(\int x(x^{2}+2)^{3}dx\).
We let \(u=x^{2}+2\) because the derivative \(\dfrac{du}{dx}=2x\) and \(x=\dfrac{1}{2}\dfrac{du}{dx}\). Then, \((x^{2}+2)^{3}=u^{3}=f(u)\) and the integral becomes:
\[\begin{align*} \int x(x^{2}+2)dx & = \int\frac{1}{2}\frac{du}{dx}u^{3}dx\\
& = \frac{1}{2}\int u^{3}\frac{du}{dx}dx\\
& = \frac{1}{2}\int u^{3}du\\
& = \frac{1}{2}\frac{1}{4}u^{4}+c,\,c\in\mathbb{R}\\
& = \frac{1}{8}(x^{2}+4)^{4}+c
\end{align*}\]
Find \(\int x^{2}e^{4-x^{3}}dx\).
We let \(u=4-x^{3}\) because \(\dfrac{du}{dx}=-3x^{2}\) which is \(-3\) times the other part of the integrand. So, \(x^{2}=-\dfrac{1}{3}\dfrac{du}{dx}\).
Substituting, the integral is:
\[\begin{align*} \int x^{2}e^{4-x^{3}}dx & = \int\left(-\frac{1}{3}\right)\frac{du}{dx}e^{u}dx\\
& =-\frac{1}{3}\int e^{u}\frac{du}{dx}dx\\
& =-\frac{1}{3}\int e^{u}du\\
& =-\frac{1}{3}e^{u}+c,\,c\in\mathbb{R}\\
& =-\frac{1}{3}e^{4-x^{3}}+c
\end{align*}\]
Find \(\int\frac{x^{4}}{1+x^{5}}dx\).
Let \(u=1+x^{5}\), then \(\dfrac{du}{dx}=5x^{4}\) and \(x^{4}=\dfrac{1}{5}\dfrac{du}{dx}\).
Substituting in the integral, we get:
\[\begin{align*} \int\frac{x^{4}}{1+x^{5}}dx & = \int\frac{1}{5}\frac{du}{dx}\frac{1}{u}dx\\
& =\frac{1}{5}\int \frac{du}{dx}\frac{1}{u}dx\\
& =\frac{1}{5}\int\frac{1}{u}du\\
& =\frac{1}{5}\ln\left|u\right|+c,\,c\in\mathbb{R}\\
& =\frac{1}{5}\ln\left|1+x^{5}\right|+c
\end{align*}\]
Substitution with definite integrals
We can also use the method of substitution to evaluate definite integrals, but we need to be careful with the limits of integration when we change the variable of integration.
For example, suppose we want to evaluate \(\int_{0}^{1}(2x+1)^{5}dx\). We let \(u=2x+1\), which means \(\dfrac{du}{dx}=2\) and so \(1=\dfrac{1}{2}\dfrac{du}{dx}\).
Note that when \(x=0\), \(u=1\) and when \(x=1\), \(u=3.\)
Substituting into the integral, we get:
\[\begin{align*} \int_{0}^{1}(2x+1)^{5}dx & =\int_{0}^{1}1\times(2x+1)^{5}dx\\
& =\int_{0}^{1}\frac{1}{2}\frac{du}{dx}\times u^{5}dx\\
& =\frac{1}{2}\int_{1}^{3}u^{5}du\\
& =\frac{1}{2}\frac{1}{6}\left[u^{6}\right]_{u=1}^{u=3}\\
& =\frac{1}{12}(3^{6}-1^{6})\\
& =\frac{1}{12}(729-1)\\
& =\frac{728}{12}\\
& =\frac{182}{3}
\end{align*}\]
In the third line of working, the limits have changed since we are integrating with respect to \(u\) instead of with respect to \(x\). This is a critical step! If you miss it, you will not get the correct answer.
