3.4 Integration by Substitution

Evaluate ${\displaystyle \int \frac{x}{\sqrt{4-x^2}}dx}$.

Answer:

This integrand is more complicated than anything in our list of basic integral formulas, so we’ll have to try something else. The only tool we have is substitution, so let’s try that!

Let $u$ be some part of the integrand. A good first choice is one step inside the messiest bit: Let $u=4-x^2$.

Compute $du=\frac{du}{dx}dx$:
$du=-2xdx$. There is $xdx$ in the integrand, so that’s a good sign; that will be $-\frac{1}{2}du$.

Translate all your $x$‘s into $u$‘s everywhere in the integral, including the $dx$:

$$\begin{array}{rl}\int \frac{x}{\sqrt{4-x^2}}dx& =\int \frac{1}{\sqrt{4-x^2}}(xdx)\\ & =\int \frac{1}{\sqrt{u}}(-\frac{1}{2}du)\\ & =-\frac{1}{2}\int \frac{1}{\sqrt{u}}du\\ & =-\frac{1}{2}\int u^{-1/2}du\end{array}$$

Integrate the new $u$-integral, if possible:
$$-\frac{1}{2}\int u^{-1/2}du=-\frac{1}{2}\frac{u^{1/2}}{1/2}+C=-u^{1/2}+C$$

Finally, substitute back $x$‘s for $u$‘s everywhere in the answer: Undoing our $u=4-x^2$ substitution yields $$-u^{1/2}+C=-\sqrt{4-x^2}+C.$$

Thus we have found $$\int \frac{x}{\sqrt{4-x^2}}dx=-\sqrt{4-x^2}+C$$

How would we check this? By differentiating:
$$\begin{array}{rl}\frac{d}{dx}(-\sqrt{4-x^2}+C)& =\frac{d}{dx}(-{(4-x^2)}^{1/2}+C)\\ & =-\frac{1}{2}{(4-x^2)}^{-1/2}(-2x)\\ & =x{(4-x^2)}^{-1/2}\\ & =\frac{x}{\sqrt{4-x^2}}\end{array}$$

Evaluate ${\displaystyle \int \frac{e^{x}dx}{{(e^x+15)}^3}}$

Answer:

This integral is not in our list of building blocks. But notice that the derivative of $e^x+15$ (which we see in the denominator) is just $e^x$ (which we see in the numerator), so substitution will be a good choice for this.

Let $u=e^x+15$. Then $du=e^{x}dx$, and this integral becomes $\int \frac{du}{u^3}=\int u^{-3}du$.

Luckily, that is on our list of building block formulas: $\int \frac{du}{u^3}=\frac{u^{-2}}{-2}+C=-\frac{1}{2u^2}+C$.

Finally, translating back: $$\int \frac{e^{x}dx}{{(e^x+15)}^3}=-\frac{1}{2{(e^x+15)}^2}+C.$$

Evaluate

a. ${\displaystyle \int \frac{x^2}{x^3+5}dx}$
b. ${\displaystyle \int \frac{x^3+5}{x^2}dx}$

Answer:

a. This is not a basic integral, but the composition is less obvious. Here, we can treat the denominator as the inside of the $\frac{1}{x}$ function.

Let $u=x^3+5$. Then $du=3x^{2}dx$. Solving for $x^{2}dx$, $x^{2}dx=\frac{du}{3}$. Substituting, $$\int \frac{x^2}{x^3+5}dx=int\frac{1}{u}\frac{du}{3}=\frac{1}{3}\int \frac{1}{u}du$$

Using our basic formulas, $$\frac{1}{3}\int \frac{1}{u}du=\frac{1}{3}\ln |u|+C.$$

Undoing the substitution, $$\int \frac{x^2}{x^3+5}dx=\frac{1}{3}\ln |x^3+5|+C.$$

b. It is tempting to start this problem the same way we did the last, but if we try it will not work, since the numerator of this fraction is not the derivative of the denominator. Instead, we need to try a different approach. For this problem, we can use some basic algebra:
$$\begin{array}{rl}\int \frac{x^3+5}{x^2}dx& =\int (\frac{x^3}{x^2}+\frac{5}{x^2})dx\\ & =\int (x+5x^{-2})dx.\end{array}$$

We can integrate this using our basic rules, without needing substitution:
$$\begin{array}{rl}\int (x+5x^{-2})dx& =\frac{x^2}{2}+5\frac{x^{-1}}{-1}+C\\ & =\frac{1}{2}{x}^2-\frac{5}{x}+C.\end{array}$$

Evaluate $\underset{0}{\overset{1}{\int }}(3x-1{)}^{4}dx$

Answer:

We’ll need substitution to find an antiderivative, so we’ll need to handle the limits of integration carefully. Let’s solve this example both ways.

Doing the antiderivative as a side problem:

Step One – find the antiderivative, using substitution: Let $u=3x-1$. Then $du=3dx$ and $$\int (3x-1{)}^{4}dx=\int u^4\left(\frac{1}{3}du\right)=\frac{1}{3}\frac{u^5}{5}+C.$$

Translating back to x: $$\frac{1}{3}\frac{u^5}{5}+C=\frac{(3x-1{)}^5}{15}+C.$$

Step Two – evaluate the definite integral: $$\underset{0}{\overset{1}{\int }}(3x-1{)}^{4}dx={\frac{(3x-1{)}^5}{15}|}_0^1=\frac{{(3(1)-1)}^5}{15}–\frac{{(3(0)-1)}^5}{15}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}.$$

Substituting for the limits of integration: Let $u=3x-1$. Then $du=3dx$ and, substituting for the limits of integration, when $x=0$, $u=-1$, and when $x=1$, $u=2$. So,
$$\begin{array}{rl}{\int }_{x=0}^{x=1}(3x-1{)}^{4}dx& ={\int }_{u=-1}^{u=2}{u}^4\left(\frac{1}{3}du\right)\\ & ={\frac{u^5}{15}|}_{u=-1}^{u=2}\\ & =\frac{(2{)}^5}{15}-\frac{(-1{)}^5}{15}\\ & =\frac{32}{15}-\frac{-1}{15}\\ & =\frac{33}{15}\end{array}$$

Evaluate $\underset{2}{\overset{10}{\int }}\frac{{(\ln \left(x\right))}^6}{x}dx$.

Answer:

I can see the derivative of $\ln \left(x\right)$ in the integrand, so I can tell that substitution is a good choice.

Let $u=\ln \left(x\right)$. Then $du=\frac{1}{x}dx$. When $x=2$, $u=\ln \left(2\right)$. When $x=10$, $u=\ln \left(10\right)$. So the new definite integral is
$$\begin{array}{rl}\underset{x=2}{\overset{x=10}{\int }}\frac{{(\ln \left(x\right))}^6}{x}dx& =\underset{u=\ln \left(2\right)}{\overset{u=\ln \left(10\right)}{\int }}{u}^{6}du\\ & ={\frac{u^7}{7}|}_{u=\ln \left(2\right)}^{u=\ln \left(10\right)}\\ & =\frac{1}{7}({(\ln \left(10\right))}^7-{(\ln \left(2\right))}^7)\\ \approx & 49.01.\end{array}$$