Calculus 2 Lesson 1: Substitution

1. Integration by Substitution
   1. Reverse Chain Rule
   2. Example
   3. Exercise
   4. Definite Integrals
   5. Method 1
   6. Method 2: Update bounds
   7. Exercises
2. Classwork
3. Homeworks

Integration by Substitution

The notes below are a very brief review. If you have never encountered integration by substitution before, you may wish to look at the following:

• Fall 2022, Calculus 1, Lesson 22
• Fall 2022, Calculus 1, Lesson 23

Those notes come with embedded videos where I go over some problems and provide more depth in my explanations. There are also lots of videos on YouTube that go over this technique. Here are some examples:

1. Khan Academy
2. Khan Academy Example 2
3. PatrickJMT
4. PatrickJMT more complicated examples

Recall: chain rule

• Suppose $h(x) = f(g(x))$.
• Let $u = g(x)$. Then:
• $h^\prime(x) = f^\prime(g(x)) g^\prime(x)$.
• Or: $h^\prime(x) = f^\prime(u) u^\prime$.
• Or: $\frac{dh}{dx} = \frac{dh}{du} \frac{du}{dx}$

Reverse Chain Rule

So let’s do it in reverse!

• $\int f^\prime(g(x)) g^\prime(x) dx$.
• Let $u = g(x)$.
• $\frac{du}{dx} = g^\prime(x)$.
• $\int f^\prime(u) \frac{du}{dx} dx$.
• $\int f^\prime(u) du = f(u) + C$.

Example

Consider $\int 2x e^{x^2} dx$.

What would be an appropriate substitution?

• $u = x^2$
• $du = 2x dx$
• $\int e^u du = e^u + C = e^{x^2} + C$

For u: pick a term whose derivative will cancel out.

Exercise

For the following exercises, make an appropriate u-substitution and compute the general antiderivative:

1. $\int (\sin(x))^2 \cos(x) dx$
2. $\int \sin(x) (\cos(x))^2 dx$
3. $\int \frac{\sin(x)}{\cos(x)} dx$
4. $\int \frac{\sin(x)}{(\cos(x))^2} dx$

Solutions

1. $\int (\sin(x))^2 \cos(x) dx = \frac{(\sin(x))^3}{3} + C$
2. $\int \sin(x) (\cos(x))^2 dx = -\frac{(\cos(x))^3}{3} + C$
3. $\int \frac{\sin(x)}{\cos(x)} dx = -\ln|\cos(x)| + C$, or $\ln|\sec(x)| + C$.
4. $\int \frac{\sin(x)}{(\cos(x))^2} dx = \sec(x) + C$.

Definite Integrals

Two ways to use substitution for definite integrals:

1. Convert back to x. Ignore the bounds until the very end. Substitute as usual, integrate, and re-introduce the bounds at the end when you have found an antiderivative.
2. Update the bounds first.

Method 1

Consider: $\int_0^1 x \sqrt{x^2 + 1} dx$.

\[\begin{align} u = x^2 + 1 \\ du = 2x dx \\ \int \frac{1}{2} \sqrt{u} du = \frac{u^{3/2}}{3} \\ = \frac{(x^2 + 1)^{3/2}}{3} \end{align}\]

Now bring back the bounds.

\[\begin{align} \int_0^1 x \sqrt{x^2 + 1} dx \\ = \left.\frac{(x^2 + 1)^{3/2}}{3}\right|_0^1 \\ = \frac{2 \sqrt{2}}{3} - \frac{1}{3} \end{align}\]

Method 2: Update bounds

Same problem:

\[\begin{align} u = x^2 + 1 \\ u(0) = 0^2 + 1 = 1 \\ u(1) = 1^2 + 1 = 2 \\ \int_1^2 \frac{1}{2} \sqrt{u} du \\ = \left.\frac{u^{3/2}}{3} \right|_1^2 = \frac{2^{3/2}}{3} - \frac{1^{3/2}}{3} \end{align}\]

Exercises

Use any method:

1. $\int_0^1 \frac{x}{x^2 + 1} dx$
2. $\int_{\pi/4}^{3\pi/4} (\sin(x))^2 \cos(x) dx$
3. $\int_0^2 xe^{-x^2}dx$

Classwork

For this lesson, please do the following problems: Section 1.5: 254, 262-270, 272, 276.

Hand in on Thursday: 254, 272.

Homeworks

1. Reminder that there is a homework on BrightSpace which is due Monday, 1/26.
2. The MyOpenMath homework is due Friday, 1/30 at 11:59 PM.