Hi again everyone! As promised in Lesson 1 (see link here), you’ll be learning a whole set of techniques for integrating more and more complicated functions. You’ll see the first of these techniques today.

Lesson 3: Integration by Substitution & Integrals Involving Exponential and Logarithmic Functions

#### Learning goals:

- Use integration by substitution to evaluate definite and indefinite integrals.
- Integrate functions involving exponential and logarithmic functions.

**Topic**:

- Volume 2, Section 1.5
*Substitution*(link to textbook section) - Volume 2, Section1.6
*Integrals Involving Exponential and Logarithmic Functions*(link to textbook section)

**WeBWorK**:

- Integration – Substitution
- Integration – Exponential and Logarithmic

**Motivating question**

How do we integrate a function that is the derivative of a composition?

Differentiate: \[f(x) = \cos^{23}(x)\]

It might help to rewrite the function we’re asked to differentiate as $f(x) = (\cos(x))^{23}$. This makes it clear that $f(x)$ is a composition: *first* we take cosine of $x$ and *then* we raise that to the 23rd power.

Since $f(x)$ is a composition, we need to use the *chain rule* to differentiate it, so $f'(x) = 23 \cdot (\cos(x))^{23-1}\cdot\frac{d}{dx}(\cos(x)) = -23\cos^{22}(x)\sin(x)$.

## Indefinite integration by substitution

In the warmup exercise we saw that if $f(x) = f(x) = (\cos(x))^{23}$, then its derivative is $f'(x) =-23\cos^{22}(x)\sin(x)$. Remember that the factor of $\sin(x)$ appears because we used the chain rule:

\[\frac{d}{dx}(F(g(x)) = F'(g(x))g'(x).\]

Once again, we can rewrite this relationship in terms of derivatives, as a relationship in terms of antiderivatives using indefinite integrals, by integrating both sides with respect to $x$:

\[F(g(x)) + C = \int F'(g(x)) g'(x) dx. \]

To integrate by substitution, our job is to figure out what the functions in the composition are. The technique is often called $u$-substitution because we let $u=g(x)$ as in the following theorem:

**Integration by substitution**: Let $F(x)$ and $g(x)$ be differentiable functions where the domain of $F(x)$ contains the range of $g(x)$. If $u=g(x)$ then $du=g'(x)dx$ and

\[\int F'(g(x)) g'(x) dx = \int F(u) du = F(u) + C = F(g(x))+C.\]

Our main priority when applying this integration technique is to determine what the function $u$ should be. Notice that $u$ appears in the integrand as the “inside” function of a composition and its derivative $du$ also appears as a factor. Figuring out a good choice of $u$ takes some trial and error but it gets easier with practice. You’ll know you’ve made a good choice of $u$ when

- you can rewrite the integrand from being completely in terms of $x$ to being completely in terms of $u$ (no $x$’s and $u$’s in the same step), and
- you can integrate your rewritten integrand with respect to $u$.

**Example. **Consider the indefinite integral $\int (-23\cos^{22}(x)\sin(x)) dx$ (this should look familiar from the warmup exercise, so we already know what we want the answer to be). We notice that $\cos(x)$ is the “inside” function of a composition and $\sin(x)dx$ appears as a factor of the integrand. This tells us that setting $u = \cos(x)$ might be a good choice. We differentiate both sides of the equation $u = \cos(x)$ to see that $du = -\sin(x)dx$. We use these two equations relating $x$ and $u$ as our dictionary to translate our integral from $x$-language to $u$-language:

\[\int (-23\cos^{22}(x)\sin(x)) dx = \int 23 u^{22} du.\]

Then we can evaluate the integral with respect to $u$ and use the same dictionary to translate back from $u$-language to $x$-language:

\[\int 23 u^{22} du = u^{23} + C = \cos^{23}(x) +C.\]

#### Examples of indefinite integration using substitution

**Video 1** below shows a short lesson as well as three examples of different types.

**Videos 2 to 7** below show more examples of indefinite integration by substitution.

## Definite integration by substitution

Don’t forget:

- An indefinite integral is a general antiderivative.
- A definite integral is a signed area.
- The Fundamental Theorem of Calculus part 2 (FTC 2) relates definite integrals and indefinite integrals.

What this says you can take what you know about indefinite integration by substitution and apply it to definite integrals. You just have to be careful to deal with your upper and lower limits of integration in terms of and $x$-language and $u$-language too. **Video 8** below shows two ways of approaching this for one example.

## Integrals Involving Exponential and Logarithmic Functions

We’ve already seen a few examples involving integration of exponential and logarithmic functions. If you remember the derivatives $\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx} \ln(x) = \frac{1}{x}$ then you’ll be able to recognize the integral versions: $\int e^x dx = e^x + C$ and $\int \frac{1}{x}dx = \ln(|x|)+ C$ (don’t forget that the domain of the natural logarithm function is all positive real numbers). See **Videos 3, 4, and 5** above for examples using these functions for integration by substitution.

### Additional Video Resources

- Evaluate the indefinite integral: \[\int \frac{x}{\sqrt{x^2+9}}dx.\]
- Evaluate the definite integral: \[\int_0^1 \frac{x}{\sqrt{x^2+9}}dx.\]