Welcome back everyone! So far you’ve learned about direct integration, integration by substitution, and integration by parts. Here is the next integration technique for you to add to your toolbox.

Lesson 5: Trigonometric Integrals

#### Learning goals:

- Evaluate definite and indefinite involving products and powers of sin(x) and cos(x).
- Evaluate definite and indefinite involving products and powers of tan(x) and sec(x).

**Topic**:

- Volume 2, Section 3.2
*Trigonometric Integrals*(link to textbook section)

**WeBWorK**:

- Integration – Trigonometric Integrals

**Motivating question**

How can we integrate products of powers of trigonometric functions?

#### Warmup exercise 1

Differentiate:

\[f(x) = \sin^3(x)\]

#### Answer 1

Since $f(x)$ is a composition, we’ll use the chain rule to find its derivative:

\[f'(x) = 3 \sin^{3-1}(x) \cdot \frac{d}{dx}(\sin(x)) = 3 \sin^2(x)\cos(x).\]

## Trigonometric Integrals

Today’s integration technique is not actually new. It will just combine things that you already know but probably haven’t used together yet: either trigonometric identities + integration by substitution or trigonometric identities + integration by parts.

#### Preliminary example

Once again, we can rewrite the relationship between $f(x)=\sin^3(x)$ and its derivative $f'(x)=3\sin^2(x)\cos(x)$ as an equation with integrals:

\[\int 3 \sin^2(x) \cos(x) dx = \sin^3(x) + C.\]

Notice, even if we hadn’t completed the warmup exercise, we could have evaluated this integral using substitution:

Let $u = \sin(x)$. Then $du = \cos(x)$. So:

\[\int 3\sin^2(x) \cos(x) dx = \int 3 u^2 du = u^3 + C = \sin^3(x)+C.\]

(We know that this is the correct indefinite integral because we already differentiated the answer in the warmup exercise and got the integrand!)

#### Motivating example

The next exercise will be similar, but here’s a hint: which trigonometric identity will help you rewrite the integrand to make it look a bit more like the preliminary example? Try it for yourself before clicking on the answer.

#### Exercise 2

Evaluate:

\[\int \sin^3(x) \cos^3(x)dx.\]

#### Answer 2

We’d like to integrate by substitution, but it’s not clear yet what $u$ should be. First, we’ll rewrite the integrand using the Pythagorean trigonometric identity $\sin^2(x) + \cos^2(x) = 1$, so we can replace $\cos^2(x)$ by $1 – \sin^2(x)$ before we make our substitution:

\[\int \sin^2(x) \cos^3(x)dx = \int \sin^2(x)\cos^2(x)\cos(x)dx\]

\[= \int \sin^2(x)(1 – \sin^2(x))\cos(x)dx.\]

We have a few choices for how to proceed next. Let’s try simplifying the radicand so we can see more easily which integration technique from our toolbox will be most appropriate:

\[\int \sin^2(x)(1 – \sin^2(x))\cos(x)dx = \int (\sin^2(x)\cos(x) – \sin^4(x)\cos(x))dx\]

\[= \int \sin^2(x)\cos(x) dx – \int \sin^4(x) \cos(x)dx.\]

Now we can see that there are two integrals to evaluate and they can both be evaluated using substitution where $u = \sin(x)$ and $du= \cos(x)$. The first term is *almost* the integral we evaluated in the previous example; the second term is *almost* identical:

\[\int \sin^2(x)\cos(x) dx – \int \sin^4(x) \cos(x)dx = \int u^2 du – \int u^4 du.\]

Now we are ready to integrate with respect to $u$ and substitute back in terms of $x$:

\[\int u^2 du – \int u^4 du = \frac{u^3}{3} – \frac{u^5}{5}+C\]

\[= \frac{\sin^3(x)}{3} – \frac{\sin^5(x)}{5}+C.\]

## More examples

Exercise 2 is a straightforward example showing how to combine trigonometric identities and substitution to evaluate integrals. Noice that having an *odd* power of $\cos(x)$ was what made the substitution work, so we still had a factor of $\cos(x)$ left over to play the role of $du$. (It’s convenient that $\sin(x)$ and $\cos(x)$ are each other’s derivatives…up to a possible negative sign.) We can make more complicated examples by using other trigonometric functions and other powers of them. You’ll have to remember your trig identities as well as the derivatives of the other four trigonometric functions.

**Video 1** shows an example that’s similar to ones we’ve seen already, but where $x$ has a coefficient.

**Video 2** below shows how to evaluate $\int \sin^2(x) dx$ (with no $\cos(x)$ factor).

**Video 3 **shows an example involving $\tan(x)$ and $\sec(x)$ instead of $\sin(x)$ and $\cos(x)$.

While the techniques in the previous examples were all rather similar, **Video 4** shows an example trigonometric identities are used together with another integration technique that you’ve seen before.

Your textbook has a list of reduction formulas to help you reduce the powers of the trigonometric factors of your integrand. The proofs of these formulas are similar to the examples you’ve seen above. Your professor may want you to understand where these formulas came from or they may want you to memorize them. Make sure to ask your professor about these formulas.

#### A remark about definite integrals

As usual, we’ve focused on indefinite integrals in this lesson. The techniques can always be combined with the Fundamental Theorem of Calculus (part 2) from Lesson 2 (link here). Don’t forget that when you are integrating by substitution, you have to deal with the limits of integration in some way. You can re-watch Video 8 from Lesson 3 (link here) for a refresher.

## A remark about integration technique names (don’t get it twisted!)

Notice that most of this lesson is about using identities for **trigonometric** functions in order to perform integration by **substitution**. But the title of this Lesson is ** not** “trigonometric substitution”. The reason for that is that there is a

**technique called “trigonometric substitution,” which we will see in the next two lessons.**

*different*### Additional Video Resources

#### Exit Ticket

Evaluate:

\[ \int \sin^3(x)\cos^8(x)dx\]