• Use pattern recognition to find an indefinite integral.
  • Use a change of variables to find an indefinite integral.
  • Use the General Power Rule for Integration to find an indefinite integral.
  • Use a change of variables to evaluate a definite integral.
  • Evaluate a definite integral involving an even or odd function.

Assignment

  • Vocabulary and teal boxes
  • p343 1–4, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 59, 63, 67–79 odd, 82, 86 91–95 odd, 117–120

Additional Resources

Pattern Recognition

With the fundamental theorem now in our tool belt, definite integrals will be a focus for a while, and we need to move past the basic rules introduced in 4.1. The first hurdle we’ll tackle is dealing with unwrapping the chain rule.

Differentiating something like $\sin \left(x^2\right)$ requires use of the chain rule, and in this case results in $\cos \left(x^2\right) \cdot 2x$. You need to differentiate the outer function, then multiply it by the derivative of the inner function.

\[\begin{align} \frac{d}{dx} F(g(x)) = F'(g(x)) \cdot g'(x) \end{align}\]

When finding antiderivatives, that pattern on the right is something we need to look for: a composite function multiplied by the derivative of the inner function. Find it, and you can work backwards from there.

Substitution Example

Find $\displaystyle \int (x^2 + 1)^2 \cdot 2x \, dx$.

In the integral, we have composite function with $(x^2 + 1)^2$, and it’s multiplied by the inner’s derivative, $2x$. Since the pattern we’re looking for exists, we can move on, but we’re going to use substitution to make our lives easier.

First, we’re going to let $u = x^2 + 1$. By differentiating that, we get $\frac{du}{dx} = 2x$, which we can rewrite as ${du = 2x \, dx}$. Now we can substitute and rewrite our integral.

\[\begin{align} &\int (x^2 + 1)^2 \cdot 2x \, dx \\ &\int (u)^2 \, du \end{align}\]

That’s much more manageable. After antidifferentiating, we can substitute back so our antiderivative is back in terms of $x$.

\[\begin{align} \int u^2 \, du &= \frac{1}{3}u^3 + C \\ &=\frac{1}{3}(x^2+1)^3 + C \end{align}\]

$\blacksquare$

So this pattern is very specific, and obviously it won’t always be there, but we can cast a wider net by multiplying by a constant.

Multiplying by a Constant Example 1

Find $\displaystyle \int x(x^2 + 1)^2 \, dx$.

The integral above is close to what we need, but requires $2x$ where it only has an $x$. If we multiply by $2$, and by $1/2$ at the same time, we’ll get the $2$ that we need without changing the value of the integral.

\[\begin{align} \int x(x^2 + 1)^2 \, dx &= \int \frac{1}{2} \cdot 2 \cdot x(x^2 + 1)^2 \, dx \\ &= \frac{1}{2} \int u^2 \, du &&\text{Let }u=x^2 + 1, du = 2x\, dx \\ &= \frac{1}{2} \left(\frac{1}{3}u^3 + C\right) \\ &= \frac{(x^2+1)^3}{6} + C &&\text{Substitute in for } u \end{align}\]

$\blacksquare$

Note that $C$ is still just $C$ even though it was multiplied by $\frac{1}{2}$. It’s a stand-in for all constants, so whether or not it was halved is irrelevant.

Multiplying by a Constant Example 2

Find $\displaystyle \int \sin^2 3x \cos 3x \, dx$.

Here, our inner function will be $u=\sin3x$, making $du=(\cos3x) (3)\, dx$. We need to add a $3$ to our original, so that means $\frac{1}{3}$ on the outside to compensate.

\[\begin{align} \int \sin^2 3x \cos 3x \, dx &= \frac{1}{3}\int u^2\, du \\ &= \frac{1}{3}\left( \frac{u^3}{3} \right) + C \\ &= \frac{\sin^33x}{9}+C \end{align}\]

$\blacksquare$

Change of Variables for Definite Integrals

So far, we’ve dealt with indefinite integrals. Definite integrals, however, require that we adjust our interval bounds when we change our variable. When you decide what $u$ will equal, you need to run your old interval bounds though that equation.

Definite Integral Change of Variable Example

Find $\displaystyle \int_0^1 x(x^2+1)^3 \, dx$.

We’ll let $u=x^2+1$, meaning our old interval of $[0,1]$ becomes $[1,2]$ because $(0)^2 + 1 = 1$ and $(1)^2 + 1 = 2$.

\[\begin{align} \int_0^1 x(x^2+1)^3 \, dx &= \frac{1}{2}\int_0^1 2x(x^2+1)^3 \, dx \\ &= \frac{1}{2}\int_1^2 u^3 \, du &&{\text{Let }u=x^2 + 1, du = 2x\, dx}\\ &= \frac{1}{2} \left[\frac{1}{4}u^4\right]_1^2 \\ &= \frac{1}{2} \left(4 - \frac{1}{4}\right) \\ &= \frac{15}{8} \end{align}\]

$\blacksquare$

Note that we don’t need to substitute our equation back in for $u$ since we already adjusted the interval. Changing the variable to $u$ doesn’t change the result of the integral, as long as the interval is adjusted properly.

Alternate Example 9

Evaluate $ \displaystyle \int_1^5 \frac{x}{\sqrt{2x-1}} \, dx$.

The book evaluates example 9 one way, but below is a method more in line with what we’ve done so far. Both versions introduce how to handle extra $x$ terms after you perform your substitution. I still recommend looking through the book’s version, as their way may be helpful in some problems.

\[\begin{align} \int_1^5 \frac{x}{\sqrt{2x-1}} \, dx &= \frac{1}{2} \int_1^5 2x(2x-1)^{-1/2} \, dx && \text{In anticipation of }u=2x - 1, du = 2\, dx \\ &= \frac{1}{2} \int_1^9 \frac{u-1}{2}u^{-1/2} \, du && \text{Since } u=2x - 1, x=\frac{u-1}{2} \\ &= \frac{1}{4} \int_1^9 u^{1/2} + u^{-1/2} \, du \\ &= \frac{1}{4} \left[\frac{2}{3}u^{3/2} + 2u^{1/2}\right]_1^9 \\ &= \frac{1}{4}\left(24-\frac{8}{3}\right) \\ &= \frac{16}{3} \end{align}\]

$\blacksquare$

The Rest

Two parts of this section deal with shortcuts that can make your life easier.

The General Power Rule for Integration

You’ll likely pick up on this pattern as you complete problems involving integrating powers. Use it as you see fit.

Integration of Even and Odd Functions

The even and odd shortcut has you taking advantage of symmetry in certain functions. Even functions are mirrored over the $y$-axis, so you can just integrate one side and double it. Odd functions have symmetry with respect to the origin, meaning the areas end up cancelling each other out. Just watch your intervals since they need to be symmetrical in both cases.