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Integration by Substitution

Definition

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.

Common Substitutions

Some examples of common substitutions.

Note: This is not a comprehensive list of possible substitutions.

Linear Substitution

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}

Example: Linear expression in $x$ raised to a power $n \gt 1$

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 $x$:

\[\int(3x+2)^5\;\mathrm{d}x=\dfrac{1}{18}(3x+2)^6+\mathrm{C}.\]

Example: Trigonometric function with argument in the form $ax+b$

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 $x$:

\[\int\sin{(5x+1)}\;\mathrm{d}x=-\dfrac{1}{5}\cos{(5x+1)}+\mathrm{C}.\]

Example: Reciprocal of a linear expression $ax+b$

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 $x$:

\[\int\dfrac{1}{2x-7}\;\mathrm{d}x=\dfrac{1}{2}\ln{\vert 2x-7 \vert}+\mathrm{C}.\]

Integrating the product of a function with the derivative of its argument

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:

\[\int f\bigl(g(x)\bigl)g'(x)\;\mathrm{d}x=\int f(u)\;\mathrm{d}u.\]

Example

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}

Trigonometric Substitution

In some cases it is convenient to substitute a trigonometric function of $u$ for a polynomial function of $x$.

Example: Reciprocal of a function in the form $1+a^2x^2$

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:

\[\int\frac{1}{1+a^2x^2}\mathrm{d} x = \frac{1}{a}\arctan(ax) + C.\] }}

Definite Integration by Substitution

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 $u(b)$.

Example

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,

\[\int_0^2 (x+2)^3\;\mathrm{d}x=60.\]

Worked Example

Example 1

Use a suitable substitution to find $\begin{align}\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x.\end{align}$

Solution

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:

\[\int\dfrac{x}{(1+x^2)^2}\;\mathrm{d}x=-\dfrac{1}{2(1+x^2)}+C.\]

Video Examples

Example 1

Prof. Robin Johnson uses integration by substitution to find $\begin{align}\int x^3\sin{(1+x^2)}\;\mathrm{d}x\end{align}$.

Example 2

Prof. Robin Johnson uses a trigonometric substitution to find $\begin{align}\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^2} }\end{align}$.

Example 3

Prof. Robin Johnson uses a trigonometric substitution to integrate the reciprocal of a quadratic, $\begin{align}\int\dfrac{\mathrm{d}x}{x^2+3x+\frac{5}{2} }.\end{align}$

Example 4

Prof. Robin Johnson uses a trigonometric substitution to integrate the quotient of a cubic and a quadratic, $\begin{align}\int\dfrac{x^3-2x^2-x+3}{x^2+2}\;\mathrm{d}x\end{align}$.

Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test Yourself

Test yourself: Numbas test on integration by substitution

See Also

External Resources

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