Short and Sweet Calculus

## 5.3Integration 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 $$\int(x^{2}+3)^{17}x\ dx$$.

Solution

Let $$u=x^{2}+3$$, thus $$du=2xdx,$$ or $$\frac{1}{2}du=xdx$$. Now we can rewrite the integral as \begin{aligned} \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 $$\int x^{2}\sqrt{x^{3}+1}\ dx$$.

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 $$\int b^{x}dx$$.

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 $\int e^{(\ln b)x}dx=\frac{1}{\ln b}e^{(\ln b)x}+C=\frac{1}{\ln b}\left(\underbrace{e^{\ln b}}_{b}\right)^{x}+C=\frac{1}{\ln b}b^{x}+C.$