Integration by substitution is an integration technique which involves making a substitution to simplify the integral. This method is also sometimes referred to as “$u$-substitution”, as the letter $u$ is typically used to denote the substituted value. The exact substitution depends on the form of the given integral, as some substitutions are more appropriate for certain problems than others.
When making a substitution $u=u(x)$ the differential $\mathrm{d}x$ must be transformed so that the integral can be computed solely with respect to $u$.
By the chain rule:
\[\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x.\]
The choice of substitution is not always immediately obvious. The ability to recognise an appropriate substitution comes from practising many different examples.
Some examples of common substitutions.
Note: This is not a comprehensive list of possible substitutions.
For certain types of integral it is convenient to use a linear substitution $u=ax+b$. \begin{align} \frac{\mathrm{d}u}{\mathrm{d}x} &= a, \\ \mathrm{d}u &=\dfrac{\mathrm{d}u}{\mathrm{d}x} \mathrm{d}x = a\cdot\mathrm{d}x \Rightarrow \mathrm{d}x = \dfrac{1}{a}\mathrm{d}u. \end{align}
Consider the integral $\begin{align}\int (3x+2)^5 \;\mathrm{d}x.\end{align}$ Here the appropriate linear substitution is $u = 3x+2$, giving:
\[\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x=\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[3x+2\bigl]\mathrm{d}x=3\;\mathrm{d}x\Rightarrow\mathrm{d}x=\dfrac{1}{3}\mathrm{d}u.\]
Substituting the expressions $u=3x+2$ and $\mathrm{d}x=\dfrac{1}{3}\mathrm{d}u$ into the integral gives:
\begin{align} \int(3x+2)^5\;\mathrm{d}x &= \int u^5\cdot\dfrac{1}{3}\;\mathrm{d}u \\ &=\dfrac{1}{3}\int u^5\;\mathrm{d}u. \end{align}
This can be evaluated directly using standard integrals:
\begin{align} \dfrac{1}{3}\int u^5\;\mathrm{d}u &= \dfrac{1}{3}\cdot\dfrac{u^6}{6}+\mathrm{C} \\ &=\dfrac{u^6}{18}+\mathrm{C}. \end{align}
Substituting $u=3x+2$ gives the value of the integral in terms of the original variable $x$:
\[\int(3x+2)^5\;\mathrm{d}x=\dfrac{1}{18}(3x+2)^6+\mathrm{C}.\]
Consider the integral $\begin{align}\int\sin{(5x+1)}\;\mathrm{d}x.\end{align}$ Here the appropriate linear substitution is $u = 5x+1$, giving
\[\mathrm{d}u\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x=\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[5x+1\bigl]\mathrm{d}x=5\;\mathrm{d}x\Rightarrow\mathrm{d}x=\dfrac{1}{5}\mathrm{d}u.\]
Substituting the expressions $u=5x+1$ and $\mathrm{d}x=\dfrac{1}{5}\;\mathrm{d}u$ into the integral gives:
\begin{align} \int\sin{(5x+1)}\;\mathrm{d}x &= \int\sin{u}\cdot\dfrac{1}{5}\;\mathrm{d}u \\ &=\dfrac{1}{5}\int\sin{u}\;\mathrm{d}u. \end{align}
This can be evaluated directly using standard integrals:
\[\dfrac{1}{5}\int\sin{u}\;\mathrm{d}u = -\dfrac{1}{5}\cos{u}+\mathrm{C}.\]
Substituting $u=5x+1$ gives the value of the integral in terms of the original variable $x$:
\[\int\sin{(5x+1)}\;\mathrm{d}x=-\dfrac{1}{5}\cos{(5x+1)}+\mathrm{C}.\]
Consider the integral $\begin{align}\int\dfrac{1}{2x-7}\;\mathrm{d}x.\end{align}$ Here the appropriate linear substitution is $u=2x-7$, giving
\[\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x=\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[2x-7\bigl]\mathrm{d}x=2\;\mathrm{d}x\Rightarrow\mathrm{d}x=\dfrac{1}{2}\mathrm{d}u.\]
Substituting the expressions $u=2x-7$ and $\mathrm{d}x=\dfrac{1}{2}\mathrm{d}u$ into the integral gives:
\begin{align} \int\dfrac{1}{2x-7}\;\mathrm{d}x &= \int\dfrac{1}{u}\cdot\dfrac{1}{2}\mathrm{d}u \\ &= \dfrac{1}{2}\int\dfrac{1}{u}\;\mathrm{d}u. \end{align}
This can be evaluated directly using standard integrals:
\[\dfrac{1}{2}\int\dfrac{1}{u}\;\mathrm{d}u=\dfrac{1}{2}\ln{\vert u \vert}+\mathrm{C}.\]
Substituting $u=2x-7$ gives the value of the integral in terms of the original variable $x$:
\[\int\dfrac{1}{2x-7}\;\mathrm{d}x=\dfrac{1}{2}\ln{\vert 2x-7 \vert}+\mathrm{C}.\]
Consider the integral $\begin{align}\int f\bigl(g(x)\bigl)g'(x)\;\mathrm{d}x.\end{align}$ Let $u=g(x)$. Then $f\bigl(g(x)\bigl)=f(u)$ and:
\[\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x=\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[g(x)\Bigl]\mathrm{d}x=g'(x)\,\mathrm{d}x.\]
Substituting the expressions $f\bigl(g(x)\bigl)=f(u)$ and $g'(x)\,\mathrm{d}x=\mathrm{d}u$ into the integral gives:
\[\int f\bigl(g(x)\bigl)g'(x)\;\mathrm{d}x=\int f(u)\;\mathrm{d}u.\]
Consider the integral $\displaystyle{ \int \dfrac{4x}{\sqrt{2x^2+1} }\;\mathrm{d}x }$. Note that the derivative of $2x^2+1$ is $4x$, which appears as a factor in the integrand. The integral can be written in the form $\displaystyle\int f\bigl(g(x)\bigl)g'(x)\mathrm{d} x$ by setting $g(x)=2x^2+1$. Then $g'(x)=4x$, and the integral becomes:
\[\int\dfrac{4x}{\sqrt{2x^2+1} }\mathrm{d} x=\int\frac{1}{\sqrt{g(x)} }g'(x)\mathrm{d} x.\]
Setting $f(g)=\dfrac{1}{\sqrt{g} }$ gives:
\[\int\frac{1}{\sqrt{g(x)} }g'(x)\mathrm{d} x=\int f\bigl(g(x)\bigl)g'(x)\mathrm{d} x.\]
By the result in the section above, the integral can therefore by transformed into an integral involving only $u$ by making the substitution $u=g(x)=2x^2+1$:
\begin{align} \int\dfrac{4x}{\sqrt{2x^2+1} }\mathrm{d} x &= \int f(u)\mathrm{d} u \\ &= \int\dfrac{1}{\sqrt{u} }\mathrm{d} u \\ &= \int u^{-\frac{1}{2} }\mathrm{d} u. \end{align}
This can be evaluated directly using standard integrals:
\[\int u^{-\frac{1}{2} }\mathrm{d} u=2u^{\frac{1}{2} }+\mathrm{C}.\]
Substituting $u=2x^2+1$ gives the value of the integral in terms of the original variable $x$:
\begin{align} \int \dfrac{4x}{\sqrt{2x^2+1} } \; \mathrm{d}x &= 2(2x^2+1)^{\frac{1}{2} }+\mathrm{C} \\ &=2\sqrt{2x^2+1}+\mathrm{C}. \end{align}
In some cases it is convenient to substitute a trigonometric function of $u$ for a polynomial function of $x$.
The integral $\displaystyle\int\frac{1}{1+a^2x^2}\mathrm{d} x$, for any real number $a\neq0$, can be computed by making the substitution:
\[x=\frac{1}{a}\tan{u}.\]
Then,
\[\mathrm{d} x = \frac{\mathrm{d} x}{\mathrm{d} u}\mathrm{d} u = \frac{\mathrm{d{\mathrm{d} u}\left[\frac{1}{a}\tan{u}\right]\mathrm{d} u = \frac{1}{a} \sec^2{u} \mathrm{d} u.\]
Substituting $\displaystyle x=\frac{1}{a}\tan{u}$ and $\displaystyle \mathrm{d} x = \frac{1}{a} \sec^2{u}\mathrm{d} u$ into the integral gives:
\begin{align} \int\frac{1}{1+a^2x^2}\mathrm{d} x &= \int\frac{1}{1+a^2\left(\frac{1}{a}\tan{u}\right)^2 }\cdot\frac{1}{a}\sec^2{u}\mathrm{d} u \\ &= \frac{1}{a}\int\frac{\sec^2{u} }{1+a^2\frac{1}{a^2}\tan^2{u} }\mathrm{d} u \\ &=\frac{1}{a}\int\frac{\sec^2{u} }{1+\tan^2{u} }\mathrm{d} u. \end{align}
Note that $\dfrac{1}{a}$ is a constant so can be pulled outside the integral, and that the terms involving $a^2$ appearing within the denominator cancel out.
Recall the trigonometric identity $\sec^2{u}=1+\tan^2{u}$. Applying this identity to the integrand gives an integral which can be computed immediately:
\begin{align} \frac{1}{a}\int\frac{\sec^2{u} }{1+\tan^2{u} }\mathrm{d} u &= \frac{1}{a}\int\frac{\sec^2{u} }{\sec^2{u} }\mathrm{d} u \\ &=\frac{1}{a}\int 1 \mathrm{d} u \\ &= \frac{u}{a} + C .\end{align}
This now needs to be transformed back into the original variable $x$. The original substitution was $\displaystyle x=\frac{1}{a}\tan{u}$. Therefore,
\begin{align} \frac{1}{a} \tan u &= x\\ \\ \tan u &= ax \\ u &= \arctan(ax). \end{align}
Substituting this back into the above result gives:
\[\int\frac{1}{1+a^2x^2}\mathrm{d} x = \frac{1}{a}\arctan(ax) + C.\] }}
When using a substitution to transform a definite integral from an integral in $x$, with lower limit $a$ and upper limit $b$, to an integral in $u=u(x)$ it is essential to also transform the limits $a$ and $b$. The lower limit of the integral in $u$ is given by the value of $u(a)$, and the upper limit is given by the value of $u(b)$.
Consider the integral $\begin{align}\int_0^2 (x+2)^3\;\mathrm{d}x.\end{align}$ Let $u=u(x)=x+2$, then
\[\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x=1\cdot\mathrm{d}x\Rightarrow\mathrm{d}x=\mathrm{d}u.\]
Substituting the expressions $u=x+2$ and $\mathrm{d}u=\mathrm{d}x$ into the integral gives
\[\int_0^2 (x+2)^3\;\mathrm{d}x=\int_{u(0)}^{u(2)} u^3\;\mathrm{d}u,\]
Note: When using a substitution to transform an integral from one variable to another the limits of the integration also need to be changed.
The lower limit of the integration in $u$ is given by the value of $u$ at the lower limit of the integral in $x$. Here the lower limit of the integral in $x$ is $0$, so the lower limit of the integral in $u$ is
\[u(0)=0+2=2.\]
Similarly, the upper limit of the integration in $u$ is given by the value of $u$ at the upper limit of the integral in $x$. Here the upper limit of the integral in $x$ is $2$, so the upper limit of the integral in $u$ is:
\[u(2)=2+2=4.\]
The transformed integral is therefore:
\[\int_0^2 (x+2)^3\;\mathrm{d}x=\int_2^4 u^3\;\mathrm{d}u.\]
The value of this integral can now be found directly, without having to transform back into an expression involving the original variable $x$:
\begin{align} \int_2^4 u^3\;\mathrm{d}u &= \left[\dfrac{u^4}{4}\right]_2^4 \\ &=\dfrac{4^4}{4}-\dfrac{2^4}{4} \\ &=4^3-4 \\ &=64-4 \\ &= 60. \end{align}
Hence,
\[\int_0^2 (x+2)^3\;\mathrm{d}x=60.\]
Use a suitable substitution to find $\begin{align}\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x.\end{align}$
Let $u=1+x^2$. Observe that this substitution simplifies the denominator of the integrand;
\[\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x=\int\dfrac{x}{u^2}\;\mathrm{d}x\]
To allow integration with respect to $u$ an appropriate substitution must be made for $x\;\mathrm{d}x$, so that $x$ no longer appears in the integrand.
Recall that $\mathrm{d}u=\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x.$
Here $u=1+x^2$, so $\dfrac{\mathrm{d}u}{\mathrm{d}x}=2x$ and $\mathrm{d}u=2x\,\mathrm{d}x.$
Dividing each side by $2$ gives $x\;\mathrm{d}x=\dfrac{1}{2}\;\mathrm{d}u$, and the integral becomes:
\[\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x=\int\dfrac{1}{u^2}\cdot\dfrac{1}{2}\;\mathrm{d}u.\]
The factor of $\dfrac{1}{2}$ is a constant, so can be taken outside the integral sign:
\[\int\dfrac{1}{u^2}\cdot\dfrac{1}{2}\;\mathrm{d}u=\dfrac{1}{2}\int\dfrac{1}{u^2}\;\mathrm{d}u.\]
This integral can be evaluated directly:
\[\dfrac{1}{2}\int\dfrac{1}{u^2}\;\mathrm{d}u=\dfrac{1}{2}\Bigl(-\dfrac{1}{u}\Bigl)=-\dfrac{1}{2u} + C ,\]
where $C$ is the constant of integration.
The solution can be expressed in terms of the original variable $x$ by substituting $u=1+x^2$ into the expression:
\[\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x=-\dfrac{1}{2(1+x^2)}+C.\]
Prof. Robin Johnson uses integration by substitution to find $\begin{align}\int x^3\sin{(1+x^2)}\;\mathrm{d}x\end{align}$.
Prof. Robin Johnson uses a trigonometric substitution to find $\begin{align}\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^2} }\end{align}$.
Prof. Robin Johnson uses a trigonometric substitution to integrate the reciprocal of a quadratic, $\begin{align}\int\dfrac{\mathrm{d}x}{x^2+3x+\frac{5}{2} }.\end{align}$
Prof. Robin Johnson uses a trigonometric substitution to integrate the quotient of a cubic and a quadratic, $\begin{align}\int\dfrac{x^3-2x^2-x+3}{x^2+2}\;\mathrm{d}x\end{align}$.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test yourself: Numbas test on integration by substitution