7.4 Integration by substitution

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

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

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.

(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 \[\int\sin(2x-\pi/4)\ dx=-\frac{1}{2}\cos(2x-\pi/4)+C\]

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

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 \[\int\tan x\ dx=\ln|\sec x|+C.\]

\[\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a^{2}}\int\frac{1}{1+\left(\frac{x}{a}\right)^{2}}dx.\] 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.

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 \[\begin{align*} \int\frac{dx}{(x+2)^{2}+1} & =\int\frac{du}{u^{2}+4}\\ & =\frac{1}{2}\arctan\frac{u}{2}+C\tag{from previous example}\\ & =\frac{1}{2}\arctan\frac{x+2}{2}+C. \end{align*}\]

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 \[\begin{aligned} \int a^{x}dx & =\int e^{u}\frac{du}{\ln a}\\ & =\frac{1}{\ln a}e^{u}+C\\ & =\frac{1}{\ln a}a^{x}+C.\end{aligned}\]

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

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 \[\begin{align*} \int\frac{e^{x}}{1+e^{2x}}dx & =\int\frac{du}{1+u^{2}}\\ & =\arctan u+C\\ & =\arctan e^{x}+C.\tag{${u=e^{x}}$} \end{align*}\]

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 \[\int\cosh x\ dx=\sinh x+C.\]

\[\int\tanh x\ dx=\int\frac{\sinh x}{\cosh x}dx.\] 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 \[\int\tanh x\ dx=\ln(\cosh x)+C.\]

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 \[\int\frac{\sqrt[x]{7}}{x^{2}}dx=-\int7^{u}du=-\frac{1}{\ln7}7^{u}+C=-\frac{1}{\ln7}\sqrt[x]{7}+C.\]

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

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

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