# Integration by Substitution

## Contents

### Integration by Substitution

There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.

what the theorem looks like is this

$\int _{a}^{b}f(x)\operatorname {d} x=\int _{\alpha }^{\beta }f(g(u))g\prime (u)\operatorname {d} u$ In order to get $\alpha$ you must plug a into the function g and to get $\beta$ you must plug b into the function g.

The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section

#### Steps

 $\int \limits _{x=a}^{x=b}f(x)dx$ $=\int \limits _{x=a}^{x=b}f(x)\ {\frac {du}{du}}\ dx$ (1) i.e. ${\frac {du}{du}}=1$ $=\int \limits _{x=a}^{x=b}{\left(f(x)\ {\frac {dx}{du}}\right)\left({\frac {du}{dx}}\right)}\ dx$ (2) i.e. ${\frac {dx}{du}}\cdot {\frac {du}{dx}}=1$ $=\int \limits _{x=a}^{x=b}\left(f(x)\ {\frac {dx}{du}}\right)g'(x)\ dx$ (3) i.e. ${\frac {du}{dx}}=g'(x)$ $=\int \limits _{x=a}^{x=b}h(g(x))g'(x)dx$ (4) i.e. Now equate $\left(f(x)\ {\frac {dx}{du}}\right)$ with $h(g(x))$ $=\int \limits _{x=a}^{x=b}h(u)g'(x)dx$ (5) i.e. $g(x)=u$ $=\int \limits _{u=g(a)}^{u=g(b)}h(u)du$ (6) i.e. $du={\frac {du}{dx}}dx=g'(x)dx$ $=\int \limits _{u=c}^{u=d}h(u)du$ (7) i.e. We have achieved our desired result

### Example 1

$\int _{0}^{2}x(x^{2}+1)^{2}\operatorname {d} x$ Instead of making this a big polynomial we will just use the substitution method.

Step 1

Let $u=x^{2}+1$ Step 2

Identify $\operatorname {d} u$ $\operatorname {d} u=2x\operatorname {d} x$ Step 3

Now we plug in our limits of integration to our u to find our new limits of integration

When $x=0,u=0^{2}+1=1$ and when $x=2,u=2^{2}+1=5$ Now our integration problem looks something like this

${\frac {1}{2}}\int _{0}^{5}(x^{2}+1)^{2}(2x)\operatorname {d} x$ Step 4

When we plug in our u it looks like

${\frac {1}{2}}\int _{0}^{5}(u)^{2}\operatorname {d} u$ Step 5

Evaluate the Integral

${\frac {1}{2}}\left[{\frac {1}{3}}u^{3}\right]_{0}^{5}$ ${\frac {1}{2}}\left[\left({\frac {1}{3}}*5^{3}\right)-\left({\frac {1}{3}}*0^{3}\right)\right]$ ${\frac {1}{2}}\left[{\frac {1}{3}}*125\right]$ ${\frac {1}{2}}\left[{\frac {125}{3}}\right]$ ${\frac {125}{6}}$ As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.

### Example 2

$\int 3x^{2}(x^{3}+1)^{5}dx$ we see that $3x^{2}$ is the derivative of $x^{3}+1$ . Letting

$u=x^{3}+1$ we have

${\frac {du}{dx}}=3x^{2}$ or, in order to apply it to the integral,

$du=3x^{2}dx$ With this we may write

$\int 3x^{2}(x^{3}+1)^{5}dx=\int u^{5}du={\frac {u^{6}}{6}}+C={\frac {(x^{3}+1)^{6}}{6}}+C$ Note that it was not necessary that we had exactly the derivative of $u$ in our integrand. It would have been sufficient to have any constant multiple of the derivative.

For instance, to treat the integral

$\int x^{4}\sin(x^{5})dx$ we may let $u=x^{5}$ . Then

$du=5x^{4}dx$ and so

${\frac {du}{5}}=x^{4}dx$ the right-hand side of which is a factor of our integrand. Thus,

$\int x^{4}\sin(x^{5})dx=\int {\frac {\sin(u)}{5}}du=-{\frac {\cos(u)}{5}}+C=-{\frac {\cos(x^{5})}{5}}+C$ ## Resources

• Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution

Definite Integrals Using Substitution