Welcome back! Your integration techniques toolbox is starting to fill up, but we still have a few more techniques to put in it! Here is the next one for your toolbox. We’ll spend two lessons on this technique.

Lesson 6: Trigonometric Substitution (part 1)

#### Learning goals:

- Solve integration problems involving a sum or difference of two squares or the square root of a sum or difference of two squares.

**Topic**:

- Volume 2, Section 3.3
*Trigonometric Substitution*(link to textbook section)

**WeBWorK**:

- Integration – Inverse Trigonometric Result

**Motivating question**

How can trigonometric functions be used to evaluate non-trigonometric integrals?

Let $\sin(\theta)=\frac{x}{a}$. Determine an expression for $\cos(\theta)$ in terms of $x$ and $a$. (Hint: use the given information to label a right triangle.)

We’ll start by labeling a right triangle. We know that $\sin(\theta)$ is the length of the side opposite of $\theta$ over the length of the hypotenuse, so after we choose an angle to label by $\theta$ we label the opposite side by $x$ and the hypotenuse by $a$. Then we use the Pythagorean theorem to determine the length of the adjacent side: $\sqrt{a^2 – x^2}$. See Figure 1.

Since cosine of an angle is the length of the adjacent side over the length of the hypotenuse, $\cos(\theta) = \frac{\sqrt{a^2-x^2}}{a}$.

## Don’t be scared!

The technique of integration by trigonometric substitution can appear daunting at first; the solutions can sometimes get rather long and you might have to remember different trigonometric identities. But the main principle is what we saw in the warmup exercise…trigonometric ratios are just that: ratios. So when parts of these ratios appear in an integrand, we can replace them with the appropriate trig functions, and then exploit the trigonometric identities to get something we can integrate. So the warning is: **don’t be sacred!** Just take these one step at a time.

## Examples

The best way to see how trigonometric substitution works is through examples. **Video 1** below walks you through some of the ingredients you’ll need to remember, helps you recognize when trigonometric substitution would be an appropriate integration technique to use or if there is a more appropriate technique, and it walks you through a first straightforward example.

**Videos 2 and 3** below walk you through two more examples involving other trigonometric substitutions.

**Videos 4 to 9 **walk you through a variety of more examples using trigonometric substitution. Notice how different the actual integration steps are in each of these examples, and how this impacts what final answers look like when we substitute back using the original variable.

You have all the tools to evaluate the integrals in these videos on your own now and you need to practice using them for yourself. Pause the video after you’ve seen the question and then try to evaluate the integral on your own before playing the rest of the video.

Remember:

- if you see $a^2-x^2$, then let $x = a \sin(\theta)$,
- if you see $x^2 – a^2$, then let $x = a \sec(\theta)$,
- if you see $x^2 + a^2$ (or $a^2 + x^2$), then let $x = a \tan(\theta)$.

You’ve now seen all of the critical details of trigonometric substitution. But notice that in all of the examples, we were replacing sums and differences of squares—there was no linear $x$ term. In our next lesson, Lesson 7: Trigonometric Substitution (part 2) (link here), we’ll see how to incorporate an another algebraic technique that you’ve seen before when there is a linear $x$ term.

#### Exit Ticket

Evaluate:

$$\dfrac{1}{x^2\sqrt{36-x^2}}dx$$