Suppose we want to evaluate an integral of the form \[\int f(g(x))g'(x)dx\tag{a}\] To this end, let \(g(x)=u.\) Finding the differential of both sides, we have

\[g'(x)dx=du.\tag{b}\] If we substitute (b) into (a), we will have \[\bbox[#F2F2F2,5px,border:2px solid black]{\int f(\underbrace{g(x)}_{=u})\underbrace{g'(x)dx}_{=du}=\int f(u)du.\tag{c}}\] This equation supplies us with a method for determining integrals in a large number of cases in which the form of the integral is not obvious.

 Theorem 1. (The Substitution Rule): Suppose \(g(x)\) is a differentiable function the range of which is an interval \(I\), and \(f\) is a function defined on \(I\). Also suppose \(F(x)=\int f(x)dx\) for \(x\in I\). Then if \(u=g(x),\) we have \[\int f(g(x))g'(x)dx=\int f(u)du=F(u)+C=F(g(x))+C.\]

Show the proof …

Because \(F’=f\) it follows from the chain rule that \[\begin{aligned} \frac{d}{dx}F(g(x)) & =F'(g(x))g'(x)\\ & =f(g(x))g'(x).\end{aligned}\] It follows from the definition of antiderivative (or integral) that \[\begin{align*} \int f(g(x))g'(x)dx & =F(g(x))+C\\ & =F(u)+C\tag{${u=g(x)}$}\\ & =\int F'(u)du\\ & =\int f(u)du. \end{align*}\]


  • Because traditionally the letter \(u\) is used in the Substitution Rule, it is sometimes called \(u\)-substituiton, but instead of \(u\), we can use any letter such as \(t,w,\theta,\) etc.
Example 1

Evaluate \({\displaystyle \int(x^{2}+3)^{17}x\ dx}\).

Solution 1

Let \(u=x^{2}+3\), thus \(du=2xdx,\) or \(\frac{1}{2}du=xdx\). Now we can rewrite the integral as \[\begin{aligned} {\displaystyle \int\underbrace{(x^{2}+3)^{17}}_{u^{17}}\underbrace{xdx}_{\frac{1}{2}du}} & =\int\frac{1}{2}u^{17}du\\ & =\frac{1}{2}\frac{u^{18}}{18}+C\\ & =\frac{1}{36}(x^{2}+3)^{18}+C.\end{aligned}\]

Example 2

Evaluate \({\displaystyle \int x^{2}\sqrt{x^{3}+1}\ dx}\).

Solution 2

Let \(u=x^{3}+1\). Then \(du=3x^{2}\ dx\) or \(x^{2}dx=\frac{1}{3}du\), and \[\begin{align*} \int x^{2}\sqrt{x^{3}+1}\ dx & =\int\sqrt{\underbrace{x^{3}+1}_{=u}}\ \overbrace{x^{2}dx}^{\frac{1}{3}du}\\ & =\int\frac{1}{3}\sqrt{u}\ du\\ & =\frac{1}{3}\int u^{1/2}du\tag{${u^{1/2}=\sqrt{u}}$}\\ & =\frac{1}{3}\left(\frac{2}{3}u^{3/2}\right)+C\tag{${\int u^{\alpha}du=\frac{1}{\alpha+1}u^{\alpha+1}+C}$}\\ & =\frac{2}{9}(x^{3}+1)^{3/2}+C.\tag{${u=x^{3}+1}$} \end{align*}\]

Formula (c) may be stated as an algorithm as follows

  1. Choose \(u\), say put \(g(x)=u\).
  2. Compute \(du=g'(x)dx\).
  3. Substitute \(u\) and \(du\) in the integral. At this step, everything should be expressed in terms of \(u\) and no \(x\) should be present in the integral. If it is not possible, make another choice for \(u\).
  4. Evaluate the resulting integral.
  5. Replace \(u\) by \(g(x)\) and express the final result in terms of \(x\).
Example 3

Evaluate \[\int\sin ax\ dx,\quad\text{and}\quad\int\cos ax\ dx.\]

Solution 3

Let \(u=ax\). Then \(du=a\ dx\) and \[\begin{aligned} \int\sin ax\ dx & =\frac{1}{a}\int\sin u\ du\\ & =-\frac{1}{a}\cos u+C\\ & =-\frac{1}{a}\cos ax+C.\end{aligned}\] Similarly \[\int\cos ax\ dx=\frac{1}{a}\sin ax+C.\]

A simple substitution is so useful that is worth noting explicitly.

Theorem 2.  If \(\int f(u)du=F(u)+C\), then \[\int f(ax+b)\,dx=\frac{1}{a}F(ax+b)+C,\]where \(a\) and \(b\) are two constants.

Show the proof …

Let \(u=ax+b\). Then \(du=a\ dx\) or \(dx=\frac{1}{a}du\) and \[\begin{align*} \int f(ax+b)dx & =\int\frac{1}{a}f(u)\ du\\ & =\frac{1}{a}F(u)+C\\ & =\frac{1}{a}F(ax+b)+C\tag{${u=ax+b}$} \end{align*}\]

Example 4

Evaluate (a)\({\displaystyle \int\sec^{2}(x-1)dx}\) (b) \({\displaystyle \int\sin(2x-\pi/4)}dx\).

Solution 4

(a) Because \(\int\sec^{2}x\ dx=\tan x+C\), it follows from the above theorem that \[\int\sec^{2}(x-1)\ dx=\tan(x-1)+C.\] (b) Since \(\int\sin x\ dx=-\cos x+C\), the above theorem gives \[\int\sin(2x-\pi/4)\ dx=-\frac{1}{2}\cos(2x-\pi/4)+C\]

Example 5

Evaluate\({\displaystyle \int\cos^{2}x\ \sin x\ dx}\).


Let \(u=\cos x\). Then \[du=-\sin xdx\] and \[\begin{aligned} \int\cos^{2}x\sin xdx & =-\int u^{2}du\\ & =-\frac{u^{3}}{3}+C\\ & =-\frac{1}{3}\cos^{3}x+C.\end{aligned}\]

Example 6

Evaluate \[\int\tan x\ dx.\]


Recall that \(\tan x=\sin x/\cos x\). Let \(u=\cos x\). Then \[du=-\sin x\ dx\] and \[\begin{aligned} \int\tan x\ dx & =\int\frac{\sin x}{\cos x}dx\\ & =\int\frac{-du}{u}\\ & =-\ln|u|+C\\ & =-\ln|\cos x|+C.\end{aligned}\] Recall that \(\ln x^{\alpha}=\alpha\ln x\). So \[\begin{aligned} -\ln|\cos x|+C & =\ln\left(|\cos x|^{-1}\right)+C\\ & =\ln|\sec x|+C.\end{aligned}\] Therefore, the result can also be written as \[\int\tan x\ dx=\ln|\sec x|+C.\]

\[\bbox[#F2F2F2,5px,border:2px solid black]{\int\tan x\ dx=\ln|\sec x|+C\quad\text{or}\quad-\ln|\cos x|+C}\]

Example 7

Evaluate \[\int\frac{1}{a^{2}+x^{2}}dx.\]


\[\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a^{2}}\int\frac{1}{1+\left(\frac{x}{a}\right)^{2}}dx.\] Let \(u=x/a\). Then \[dx=a\ du\] and \[\begin{aligned} \frac{1}{a^{2}}\int\frac{1}{1+(x/a)^{2}}dx & =\frac{1}{a^{2}}\int\frac{1}{1+u^{2}}a\ du\\ & =\frac{1}{a}\arctan u+C\\ & =\frac{1}{a}\arctan\frac{x}{a}+C.\end{aligned}\] To evaluate \(\int\frac{1}{1+(x/a)^{2}}dx\), we could simply use Theorem above.

Example 8

Evaluate \[\int\frac{1}{x^{2}+2x+5}dx.\] [Hint: Complete the square in the denominator]


By completing the square in the denominator, we have \[\begin{aligned} \int\frac{1}{x^{2}+2x+5}dx & =\int\frac{dx}{(x^{2}+2x+1)+4}\\ & =\int\frac{dx}{(x+2)^{2}+2^{2}}.\end{aligned}\] Now let \(u=x+2\). Then \(du=dx\) and \[\begin{align*} \int\frac{dx}{(x+2)^{2}+1} & =\int\frac{du}{u^{2}+4}\\ & =\frac{1}{2}\arctan\frac{u}{2}+C\tag{from previous example}\\ & =\frac{1}{2}\arctan\frac{x+2}{2}+C. \end{align*}\]

Example 9

Evaluate \({\displaystyle \int a^{x}dx}\).


Let \(a^{x}=e^{u}\). Taking the natural logarithm of each side, we get \[\begin{aligned} \ln(a^{x}) & =\ln e^{u}\\ x\ln a & =u.\end{aligned}\] Thus \[x\ln a=u\Rightarrow\ln a\,dx=du.\] Now we can write \[\begin{aligned} \int a^{x}dx & =\int e^{u}\frac{du}{\ln a}\\ & =\frac{1}{\ln a}e^{u}+C\\ & =\frac{1}{\ln a}a^{x}+C.\end{aligned}\]

Example 10

Evaluate \[\int e^{ax}dx.\]


We can directly use Theorem above or use the substitution \(u=ax\). The substitution leads to \[du=a\ dx\] and \[\begin{aligned} \int e^{ax}dx & =\int e^{u}\frac{du}{a}\\ & =\frac{1}{a}\int e^{u}du\\ & =\frac{1}{a}e^{u}+C\\ & =\frac{1}{a}e^{ax}+C.\end{aligned}\]

Example 11

Evaluate \[\int\frac{e^{x}}{1+e^{2x}}dx.\]


Note that the substitution \(u=1+e^{2x}\) does not work because \(du=2e^{2x}dx(\neq e^{x}dx)\) does not appear in the numerator. But if we notice that \(1+e^{2x}=1+(e^{x})^{2}\), we can make the subtitution \(u=e^{x}\) to make the denominator of the form \(1+u^{2}\). Then \[u=e^{x}\Rightarrow du=e^{x}\ dx\] and \[\begin{align*} \int\frac{e^{x}}{1+e^{2x}}dx & =\int\frac{du}{1+u^{2}}\\ & =\arctan u+C\\ & =\arctan e^{x}+C.\tag{${u=e^{x}}$} \end{align*}\]

Example 12

Evaluate \({\displaystyle \int\sinh x\ dx}\) and \({\displaystyle \int\cosh x\ dx}\).


Because \((\cosh x)’=\sinh x\), we know that \[\int\sinh x\ dx=\cosh x+C.\] Also we can use the definition of \(\sinh x\) for evaluation of the integral: \[\sinh x=\frac{e^{x}-e^{-x}}{2}\] [Recall that \(\sinh x\) (similar to \(\sin x\)) is an odd function, so it is half of \(e^x{\color{red}-}e^{-x}\) and \(\cosh x\) (similar to \(\cos x\)) is an even function, so it is half of \(e^x{\color{red}+}e^{-x}\).] \[\begin{aligned} \int\sinh x\ dx & =\frac{1}{2}\int(e^{x}-e^{-x})\ dx\\ & =\frac{1}{2}\left[e^{x}-(-e^{-x})\right]+C\\ & =\cosh x+C\end{aligned}\] Similarly we can show \[\int\cosh x\ dx=\sinh x+C.\]

\[\bbox[#F2F2F2,5px,border:2px solid black]{\begin{array}{c} {\displaystyle \int\sinh x\ dx=\cosh x+C}\\ {\displaystyle \int\cosh x\ dx=\sinh x+C} \end{array}}\]

Example 13

Evaluate \({\displaystyle \int\tanh x\ dx}\).


\[\int\tanh x\ dx=\int\frac{\sinh x}{\cosh x}dx.\] Let \(u=\cosh x\). Then \(du=\sinh x\ dx\) and \[\begin{aligned} \int\frac{\sinh x}{\cosh x}dx & =\int\frac{du}{u}\\ & =\ln|u|+C\\ & =\ln|\cosh x|+C.\end{aligned}\] Because \(\cosh x=(e^{x}+e^{-x})/2>0\), so \(|\cosh x|=\cosh x\) and \[\int\tanh x\ dx=\ln(\cosh x)+C.\]

Example 14

Evaluate \[\int\frac{\sqrt[x]{7}}{x^{2}}dx.\]


Note that \(\sqrt[x]{7}=7^{1/x}.\) If we let \(\frac{1}{x}=u\), because \[\frac{1}{x}=u\Rightarrow-\frac{1}{x^{2}}dx=du,\] we have \[\int\frac{\sqrt[x]{7}}{x^{2}}dx=-\int7^{u}du=-\frac{1}{\ln7}7^{u}+C=-\frac{1}{\ln7}\sqrt[x]{7}+C.\]

Example 15

Evaluate \[\int\frac{dx}{x\ln x}.\]


Let \(u=\ln x\), then \[du=\frac{1}{x}dx\] and \[\begin{align*} \int\underbrace{\frac{1}{\ln x}}_{u}\underbrace{\frac{1}{x}dx}_{du} & =\int\frac{du}{u}\\ & =\ln|u|+C\\ & =\ln|\ln x|+C\tag{${u=\ln x}$} \end{align*}\]

Example 16

Evaluate \[\int\frac{\sin(\ln x)}{x}dx.\]


Let \(u=\ln x\), then \[du=\frac{1}{x}dx.\] So \[\begin{aligned} \int\frac{\sin(\ln x)}{x}dx & =\int\sin u\ du\\ & =-\cos u+C\\ & =-\cos(\ln x)+C.\end{aligned}\]

Example 17

Evaluate \[\int\frac{x}{1+x^{4}}\arctan x^{\text{2}}\ dx.\]


Let \(u=\arctan x^{2}\), then \[\begin{aligned} du & =2x\frac{1}{1+(x^{2})^{2}}dx\\ & =\frac{2x}{1+x^{4}}dx.\end{aligned}\] Therefore \[\begin{aligned} \int\frac{x}{1+x^{4}}\arctan x^{2}\ dx & =\int\frac{1}{2}\underbrace{\arctan x^{2}}_{u}\underbrace{\frac{2x}{1+x^{4}}dx}_{du}\\ & =\frac{1}{4}u^{2}+C\\ & =\frac{1}{4}\arctan^{2}x^{2}+C.\end{aligned}\]