Short and Sweet Calculus

5.3 Integration by Substitution

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 \[\boxed{\int f(\underbrace{g(x)}_{=u})\underbrace{g'(x)dx}_{=du}=\int f(u)du.\tag{c}}\] 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.

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

  • If we can detect a function and its derivative in the integrand, we may try integration by substitution.

Example 5.4. Evaluate \({\displaystyle \int(x^{2}+3)^{17}x\ dx}\).

Solution

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

Solution

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

Example 5.6. Evaluate \[\int\tan x\ dx.\]

Solution

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 \[\begin{aligned} -\ln|\cos x|+C & =\ln\left(|\cos x|^{-1}\right)+C\\ & =\ln|\sec x|+C.\tag{\ensuremath{\sec x=\frac{1}{\cos x}}}\end{aligned}\] Therefore, the result can also be written as \[\int\tan x\ dx=\ln|\sec x|+C.\] \[\boxed{\int\tan x\ dx=\ln|\sec x|+C\quad\text{or}\quad-\ln|\cos x|+C}\]

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

5.3. 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.

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

Example 5.7. Evaluate \({\displaystyle \int b^{x}dx}\).

Solution

Because \(b=e^{\ln b}\), we have \[b^{x}=\left(e^{\ln b}\right)^{x}=e^{(\ln b)x}.\tag{Recall \ensuremath{\left(A^{B}\right)^{C}=A^{BC}}}\] We know that \(\int e^{x}dx=e^{x}+C\). It follows from the above theorem that \[\int e^{(\ln b)x}dx=\frac{1}{\ln b}e^{(\ln b)x}+C=\frac{1}{\ln b}\left(\underbrace{e^{\ln b}}_{b}\right)^{x}+C=\frac{1}{\ln b}b^{x}+C.\]


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