Trigonometric Substitution

Section 7.6 Trigonometric Substitution

We’re going to look at an integral that requires a variable substitution, but our goal for the substitution will be a bit different. We’re going to focus on the structure of our integrand function, but we won’t be focusing on composition. Instead, we’re going to focus on some trigonometric identities that we’ve used already:

\begin{align*} 1-\sin^2(\theta) \amp = \cos^2(\theta)\\ \sec^2(\theta)-1 \amp = \tan^2(\theta)\\ \tan^2(\theta)+1 \amp = \sec^2(\theta) \end{align*}

🔗

Activity 7.6.1. Difference of Squares.

Consider the integral:

\begin{equation*} \int \sqrt{1-x^2}\;dx\text{.} \end{equation*}

🔗

(a)

First, convince yourself that a normal $u$-substitution will not be an effective strategy for integration in this case. Why not?

🔗

Hint.

There is composition in this integrand function, but what part of our $u$-substitution is missing?

🔗

(b)

Second, convince yourself that $\sqrt{1-x^2}\neq \sqrt{1}-\sqrt{x^2}\text{.}$ Why can we not distribute roots across sums and differences like this? When can we “distribute” roots across multiple things?

🔗

Hint.

A root is really an exponent: why can’t we distribute exponents across sums and differences? Try this with $(1-x^2)^2\text{,}$ and convince yourself that this isn’t $1^2-(x^2)^2\text{.}$

🔗

Notice, though, that something like $(4x^2)^2=4^2(x^2)^2\text{,}$ and $\sqrt{4x^2}$ could act similarly.

🔗

(c)

Our goal, then, is to utilize a substitution (using trigonometric functions) to somehow transform this difference of squared terms under the square root into a single product of squared things under the square root.

🔗

Which trigonometric identities from our list of them above utilize differences of thing squared, and equate them to a single term?

🔗

Can you use the order of the subtraction to help guide which substitution we should use?

🔗

Hint.

We want to pick some trigonometric function $T(\theta)$ so that when we let $x=T(\theta)\text{,}$ we end up with one of the Pythagorean Identities above. Is there something we can put in for $x$ such that $1-x^2$ becomes $1-(T(\theta))^2$ in a convenient way?

🔗

(d)

When we do a variable substitution in an integral, we are not only finding a way of transforming $x$ to be in terms of some other variable (in this case, $\theta$). We also need to transform the differential, $dx\text{.}$ Based on your substitution of $x=T(\theta)\text{,}$ what is $dx\text{?}$

🔗

Hint.

We’ll use $dx=T'(\theta)\;d\theta\text{.}$ What is $T'(\theta)$ for you?

🔗

(e)

Perform your substitution! Use your substitution $x=T(\theta)$ and $dx=T'(\theta)\;d\theta\text{.}$ Note that we have picked this substitution with a very specific goal: we are hoping to notice a Pythagorean Identity.

🔗

After you have performed your substitution, apply the relevant Pythagorean Identity to the radicand: the bit of our function underneath the radical or root. What integral are we left with (in terms of $\theta$)?

🔗

Hint.

Since we are using $x=\sin(\theta)\text{,}$ we should notice $1-\sin^2(\theta)$ underneath the square root.

🔗

Solution.

\begin{align*} x=\sin(\theta)\;\; \amp \;\; dx=\cos(\theta)\;d\theta\\ \int \sqrt{1-x^2}\;dx \amp = \int \sqrt{\underbrace{1-\sin^2(\theta)}_{\cos^2(\theta)}} \cos(\theta)\;d\theta\\ \amp = \int \sqrt{\cos^2(\theta)}\;\cos(\theta)\;d\theta\\ \amp = \int \cos(\theta) \cos(\theta)\;d\theta\\ \amp = \int \cos^2(\theta)\;d\theta \end{align*}

🔗

This new integral is something we can antidifferentiate now! We already have done this one in Activity 7.4.4 Squared Trig Functions. So we can end up with:

\begin{align*} \int \sqrt{1-x^2}\;dx \amp = \int \cos^2(\theta)\;d\theta\\ \amp = \frac{\theta + \sin(\theta)\cos(\theta)}{2}+C \end{align*}

It is up to us, now, to translate this antiderivative family to be written in terms of $x\text{.}$ We can utilize our substitution to do this, but let’s first think about how this variable substitution works a bit more.

🔗

Subsection Another Type of Variable Substitution

We’re going to employ another variable substitution, in the same way that we use $u$-substitution. The main difference is the goal: we’re going to select our substitution not based on uncovering the composition in our function (like in $u$-substitution). Instead, we’ll focus on selecting a trigonometric function in order to utilize the relevant Pythagorean identity to rewrite our sum or difference of squares.

🔗

$Two rectangles, one labeled "x Context" and the other, beside it, labeled "theta Context." Inside the "x Context" rectangle is an integral of f(x) dx with an arrow pointing towards F(x)+C. Inside the "theta Context" rectangle is an integral of g(theta) d theta with an arrow pointing towards G(theta)+C. There is an arrow connecting the integral of f(x) dx in the first rectangle to the integral of g(theta) d theta labeled x=T(theta) and dx = T’(theta) d theta. Then there is an arrow connecting G(theta)+C to F(x)+C labeled x = T(theta)$

Figure 7.6.1. General idea of how this variable substitution works.

🔗

Ok, but how do we choose which trigonometric function to use in our substitution? Since we’re focusing on sums or differences of squared terms, we can think of the different arrangements of terms, connect them with different Pythagorean Identities, and set up some strategies for picking a trigonometric substitution.

\begin{align*} 1-x^2 \amp \longleftrightarrow 1-\sin^2(\theta) = \cos^2(\theta)\\ x^2-1 \amp \longleftrightarrow \sec^2(\theta)-1=\tan^2(\theta)\\ x^2+1 \amp \longleftrightarrow \tan^2(\theta)+1 = \sec^2(\theta) \end{align*}

We can note that the sum is commutative, so we can treat $1+x^2$ the same way that we treat $x^2+1\text{.}$

🔗

We’ll also notice that the constant term can differ: we can scale our Pythagorean identities by some constant easily to make sure that they match.

\begin{align*} a^2-x^2 \amp \longleftrightarrow a^2-\left(a\sin(\theta)\right)^2 = \left(a\cos(\theta)\right)^2\\ x^2-a^2 \amp \longleftrightarrow \left(a\sec(\theta)\right)^2-a^2=\left(a\tan(\theta)\right)^2\\ x^2+a^2 \amp \longleftrightarrow \left(a\tan(\theta)\right)^2+a^2 = \left(a\sec(\theta)\right)^2 \end{align*}

🔗

This can be confusing, and we want to keep thinking about how we might recognize these structures to pick a substitution. Yes, we can recognize these Pythagorean identities. We can rely on the order of subtraction or noticing addition. But we also can think about this geometrically. The Pythagorean Identities come from the Pythagorean Theorem, relating the squared lengths of the sides of a right triangle together. Let’s visualize our substitutions geometrically.

🔗

We’ll consider three triangles, each with side lengths of $x$ and $a\text{.}$ The third side length will vary between $\sqrt{x^2-a^2}\text{,}$ $\sqrt{a^2-x^2}\text{,}$ and $\sqrt{x^2+a^2}$ (or the equivalent $\sqrt{a^2+x^2}$) based on which length is representing the hypotenuse.

🔗

Activity 7.6.2. Trig Substitution Geoemtry.

We’re going to consider three triangles, and we’re going to fill in side lengths. In each of these, we’ll assume that the lengths $x$ and $a$ are real numbers and are positive.

Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the hypotenuse is labeled a. — Figure 7.6.2. Three triangles to guide our trigonometric substitutions.
🔗

Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the side adjacent to theta is labeled a. — Figure 7.6.2. Three triangles to guide our trigonometric substitutions.
🔗

🔗

(a)

Use the Pythagorean theorem to label the missing side length in each of the three triangles.

🔗

(b)

For each triangle, explain how you can tell which side length represents the hypotenuse when you see the lengths $x\text{,}$ $a\text{,}$ and then the missing lengths you found above: $\sqrt{x^2-a^2}\text{,}$ $\sqrt{a^2-x^2}\text{,}$ or $\sqrt{x^2+a^2}\text{.}$

🔗

Hint.

We know that the hypotenuse is the longest side length in a triangle. Just based on the square root length, how can you tell which length is longest?

🔗

Solution.

If one of the side lengths is $\sqrt{x^2-a^2}\text{,}$ then we know that $x\gt a$ (otherwise the square root is a non-real number). We also know that $x\gt \sqrt{x^2-a^2}$ (because $a\gt 0$). This means that $x$ is the length of the hypotenuse.

🔗

If one of the side lengths is $\sqrt{a^2-x^2}\text{,}$ then we know that $a\gt x$ (otherwise the square root is a non-real number). We also know that $a\gt \sqrt{a^2-x^2}$ (because $x\gt 0$). This means that $a$ is the length of the hypotenuse.

🔗

If one of the side lengths is $\sqrt{x^2+a^2}\text{,}$ then we know that $\sqrt{x^2+a^2} \gt a$ and $\sqrt{x^2+a^2} \gt x$ (because $x,a\gt 0$). This means that $\sqrt{x^2+a^2}$ is the length of the hypotenuse.

🔗

(c)

For each triangle, find a trigonometric function of $\theta$ that connects lengths $x$ and $a$ to each other.

🔗

Solve each for $x$ to reveal the relevant substitution.

🔗

(d)

For each substitution, find the corresponding substitution for the differential, $dx\text{.}$

🔗

This gives us a nice strategy for substitution!

🔗

Trigonometric Substitution.

We have three (typical) ways of using trigonometric substitution to transform a sum or difference of squared terms into a product of squares.

For an integral containing $(a^2-x^2)\text{,}$ we can use the following triangle to build our substitution:
🔗

$Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the hypotenuse is labeled a, and the remaining side length is labeled sqrt(a^2-x^2).$

Figure 7.6.3.
🔗

This results in using the following substitution and identity.

\begin{align*} x \amp = a\sin(\theta)\\ dx \amp = a\cos(\theta)\;d\theta\\ a^2-\left(a\sin(\theta)\right)^2 \amp=\left(a\cos(\theta)\right)^2 \end{align*}

🔗

🔗
For an integral containing $(x^2-a^2)\text{,}$ we can use the following triangle to build our substitution:
🔗

$Right triangle with one of the acute angles labeled as theta. The side adjacent to theta is labeled a, and the hypotenuse is labeled x, and the remaining side length is labeled sqrt(x^2-a^2).$

Figure 7.6.4.
🔗

This results in using the following substitution and identity.

\begin{align*} x \amp = a\sec(\theta)\\ dx \amp = a\sec(\theta)\tan(\theta)\;d\theta\\ \left(a\sec(\theta)\right)^2 -a^2 \amp=\left(a\tan(\theta)\right)^2 \end{align*}

🔗

🔗
For an integral containing $(x^2+a^2)\text{,}$ we can use the following triangle to build our substitution:
🔗

$Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the side adjacent to theta is labeled a. The hypotenuse is labeled sqrt(x^2+a^2).$

Figure 7.6.5.
🔗

This results in using the following substitution and identity.

\begin{align*} x \amp = a\tan(\theta)\\ dx \amp = a\sec^2(\theta)\;d\theta\\ \left(a\tan(\theta)\right)^2+a^2\ \amp=\left(a\sec(\theta)\right)^2 \end{align*}

🔗

🔗

🔗

Two things to note before we move on:

There are really 6 main trigonometric substitutions. If you go back to Activity 7.6.2 and place the angle $\theta$ in the opposite corner of the triangle, the substitutions you build will all be using the “co-functions”: cosine, cosecant, and cotangent. Each of these has a very similar structure with regard to derivatives (for the differential substitution) and Pythagorean Identities. Each is equivalent to the respective sine, secant, and tangent substitutions. We often choose to use sine, secant, and tangent just due to familiarity.
🔗

🔗
We can use the triangle as a kind of key for our substitution! After antidifferentiating, we have some antiderivative family written in terms of an angle $\theta\text{:}$ we can use the triangle to substitution trigonometric functions of $\theta$ to be written in terms of $x\text{.}$
🔗

🔗

🔗

Example 7.6.6.

We can finish the substitution we started in Activity 7.6.1. We used the substitution $x=\sin(\theta)\text{,}$ but we now can construct the relevant triangle.

🔗

Since we were hoping so use a substitution to rewrite the difference of squares, $\sqrt{1-x^2}\text{,}$ we had the following triangle:

🔗

Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the hypotenuse is labeled 1, and the remaining side length is labeled sqrt(1-x^2). — Figure 7.6.7. Substitution used in Activity 7.6.1.
🔗

We can see that $\sin(\theta)=\dfrac{x}{1}$ or $x=\sin(\theta)\text{,}$ which was our substitution.

🔗

But we also were left with the following antiderivative:

\begin{align*} \int \sqrt{1-x^2}\;dx \amp = \int \cos^2(\theta)\;d\theta\\ \amp = \frac{\theta + \sin(\theta)\cos(\theta)}{2}+C \end{align*}

Now we can substitute that antiderivative! We can see from our triangle that $\cos(\theta)=\sqrt{1-x^2}\text{,}$ $\sin(\theta)=x$ (this also was our original substitution anyways), and we also can invert our substitution to get $\theta=\sin^{-1}(x)\text{.}$

\begin{align*} \int \sqrt{1-x^2}\;dx \amp = \int \cos^2(\theta)\;d\theta\\ \amp = \frac{\theta + \sin(\theta)\cos(\theta)}{2}+C\\ \amp = \frac{1}{2}\left(\sin^{-1}(x) + x\sqrt{1-x^2}\right)+C \end{align*}

🔗

Activity 7.6.3. Practicing Trigonometric Substitution.

Let’s look at three integrals, and practice the kind of thinking we’ll need to use to apply trigonometric substitution to them.

$\displaystyle \displaystyle \int \dfrac{\sqrt{x^2-9}}{x}\;dx$
🔗

🔗
$\displaystyle \displaystyle \int \frac{2}{(4-x^2)^{3/2}}\;dx$
🔗

🔗
$\displaystyle \displaystyle \int \frac{1}{x^2\sqrt{x^2+1}}\;dx$
🔗

🔗

For each integral, do the following:

🔗

(a)

Identify the term (or terms) that signify that trigonometric substitution might be a reasonable strategy.

🔗

Hint.

In each case, you’re looking for some sum or difference of squared terms, normally (but not always) nested inside of some square root. It also might be nice to rewrite the second integral in order to notice the root:

\begin{equation*} \int \frac{2}{(4-x^2)^{3/2}}\;dx = \int \frac{2}{\left(\sqrt{4-x^2}\right)^3}\;dx\text{.} \end{equation*}

🔗

(b)

Use that portion of the integral to compare three side lengths of a triangle. Which one is the largest (and so must represent the length of the hypotenuse)?

🔗

Hint.

For the side lengths $x\text{,}$ $3\text{,}$ and $\sqrt{x^2-9}\text{,}$ which must be the largest? Think about $\sqrt{x^2-9}$ and what we can learn from it.
🔗

🔗
For the side lengths $x\text{,}$ $2\text{,}$ and $\sqrt{4-x^2}\text{,}$ which must be the largest? Think about $\sqrt{4-x^2}$ and what we can learn from it.
🔗

🔗
For the side lengths $x\text{,}$ $1\text{,}$ and $\sqrt{x^2+1}\text{,}$ which must be the largest? Think about $\sqrt{x^2+1}$ and what we can learn from it.
🔗

🔗

🔗

Solution.

We know that $x\gt 3\text{,}$ since $x^2-9\gt 0\text{.}$ Similarly, we know that $x\gt \sqrt{x^2-9}\text{.}$ So $x$ has to be the length of the hypotenuse.
🔗

🔗
We know that $2\gt x\text{,}$ since $4-x^2\gt 0\text{.}$ Similarly, we know that $2\gt \sqrt{4-x^2}\text{.}$ So $2$ has to be the length of the hypotenuse.
🔗

🔗
We know that $\sqrt{x^2+1} \gt x\text{,}$ since we are adding $1$ to $x^2$ under the square root. This also is the reason that $\sqrt{x^2+1}\gt 1$ (we are adding $x^2$ under the root). So $\sqrt{x^2+1}$ has to be the length of the hypotenuse.
🔗

🔗

🔗

(c)

Construct the triangle, label an angle $\theta\text{,}$ and use a trigonometric function to connect the two single-term side lengths. (Feel free to change the angle you label in order to use the sine, secant, or tangent functions instead of their co-functions).

🔗

Solution.

For the integral $\displaystyle \int \frac{\sqrt{x^2-9}}{x}\;dx\text{,}$ we have the following triangle.
🔗

$Right triangle with one of the acute angles labeled as theta. The side adjacent to theta is labeled 3, and the hypotenuse is labeled x, and the remaining side length is labeled sqrt(x^2-9).$

Figure 7.6.8.
🔗
This gives us $\cos(\theta) = \dfrac{3}{x}$ or, equivalently, $\sec(\theta)=\dfrac{x}{3}\text{.}$
🔗

🔗
For the integral $\displaystyle \int \frac{2}{(4-x^2)^{3/2}}\;dx\text{,}$ we have the following triangle.
🔗

$Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the hypotenuse is labeled 2, and the remaining side length is labeled sqrt(4-x^2).$

Figure 7.6.9.
🔗
This gives us $\sin(\theta) = \dfrac{x}{2}\text{.}$
🔗

🔗
For the integral $\displaystyle \int \frac{1}{x^2\sqrt{x^2+1}}\;dx\text{,}$ we have the following triangle.
🔗

$Right triangle with one of the acute angles labeled as theta. The side opposite theta is labeled x, and the side adjacent to theta is labeled 1. The hypotenuse is labeled sqrt(x^2+1).$

Figure 7.6.10.
🔗
This gives us $\tan(\theta) = \dfrac{x}{1}$ or, equivalently, $\tan(\theta)=x\text{.}$
🔗

🔗

🔗

(d)

Define your substitution (for both $x$ and the differential $dx$), and identify the Pythagorean Identity that will be relevant for the integral.

🔗

Solution.

We will use $x=3\sec(\theta)$ and $dx=3\sec(\theta)\tan(\theta)\;d\theta\text{.}$
🔗

Then, we can expect to find the Pythagorean Identity $9\sec^2(\theta)-9=9\tan^2(\theta)\text{.}$
🔗

🔗
We will use $x=2\sin(\theta)$ and $dx=2\cos(\theta)\;d\theta\text{.}$
🔗

Then, we can expect to find the Pythagorean Identity $4-4\sin^2(\theta)=4\cos^2(\theta)\text{.}$
🔗

🔗
We will use $x=\tan(\theta)$ and $dx=\sec^2(\theta)\;d\theta\text{.}$
🔗

Then, we can expect to find the Pythagorean Identity $\tan^2(\theta)+1=\sec^2(\theta)\text{.}$
🔗

🔗

🔗

(e)

Substitute and antidifferentiate!

🔗

Hint.

$\displaystyle \int \frac{\sqrt{x^2-9}}{x}\;dx = \int \frac{\sqrt{9\sec^2(\theta)-9}}{3\sec(\theta)}\left(3\sec(\theta)\tan(\theta)\right)\;d\theta$
🔗

Also note that $\sqrt{9\sec^2(\theta)-9} = \sqrt{9\tan^2(\theta)}\text{.}$
🔗

🔗
$\displaystyle \int \frac{2}{(4-x^2)^{3/2}}\;dx = \int \frac{2}{\left(\sqrt{4-4\sin^2(\theta)}\right)^3}\left(2\cos(\theta)\right)\;d\theta$
🔗

Also note that $\sqrt{4-4\sin^2(\theta)} = \sqrt{4\cos^2(\theta)}\text{.}$
🔗

🔗
$\displaystyle \int \frac{1}{x^2\sqrt{x^2+1}}\;dx = \int \frac{1}{\tan^2(\theta)\sqrt{\tan^2(\theta)+1}}\left(\sec^2(\theta)\right)\;d\theta$
🔗

Also note that $\sqrt{\tan^2(\theta)+1} = \sqrt{\sec^2(\theta)}\text{.}$
🔗

🔗

🔗

Solution.

Using $x=3\sec(\theta)\text{,}$ we get:

\begin{align*} \int \frac{\sqrt{x^2-9}}{x}\;dx \amp = \int \frac{\sqrt{9\sec^2(\theta)-9}}{3\sec(\theta)}\left(3\sec(\theta)\tan(\theta)\right)\;d\theta\\ \amp = \int \frac{3\sqrt{9\tan^2(\theta)}\sec(\theta)\tan(\theta)}{3\sec(\theta)}\;d\theta\\ \amp = 3 \int \tan^2(\theta)\;d\theta\\ \amp 3 \int \sec^2(\theta)-1\;d\theta\\ \amp 3 \left(\tan(\theta)-\theta\right)+C\\ \amp 3\tan(\theta)-3\theta+C \end{align*}

🔗

🔗
Using $x=2\sin(\theta)\text{,}$ we get:

\begin{align*} \int \frac{2}{(4-x^2)^{3/2}}\;dx \amp= \int \frac{2}{\left(\sqrt{4-4\sin^2(\theta)}\right)^3}\left(2\cos(\theta)\right)\;d\theta \\ \amp = \int \frac{4\cos(\theta)}{\left(\sqrt{4\cos^2(\theta)}\right)^3}\;d\theta\\ \amp =\frac{1}{2}\int \frac{1}{\cos^2(\theta)}\;d\theta\\ \amp =\frac{1}{2} \int \sec^2(\theta)\;d\theta\\ \amp = \frac{1}{2}\tan(\theta)+C \end{align*}

🔗

🔗
Using $x=\tan(\theta)\text{,}$ we get:

\begin{align*} \int \frac{1}{x^2\sqrt{x^2+1}}\;dx \amp= \int \frac{1}{\tan^2(\theta)\sqrt{\tan^2(\theta)+1}}\left(\sec^2(\theta)\right)\;d\theta\\ \amp = \int \frac{\sec^2(\theta)}{\tan^2(\theta)\sqrt{\sec^2(\theta)}}\;d\theta \\ \amp = \int \frac{\sec(\theta)}{\tan^2(\theta)}\\ \amp = \int \frac{\cos(\theta)}{\sin^2(\theta)}\;d\theta\\ \amp = \int \csc(\theta)\cot(\theta)\;d\theta\\ \amp = -\csc(\theta)+C \end{align*}

🔗

🔗

🔗

(f)

Use your triangle to substitute your antiderivative back in terms of $x\text{.}$

🔗

Solution.

From the triangle, we get $\tan(\theta) = \dfrac{\sqrt{x^2-9}}{3} \text{.}$ Then, since $\sec(\theta)=\dfrac{x}{3}\text{,}$ we can use $\theta = \sec^{-1}\left(\dfrac{x}{3}\right)\text{.}$ This gives us:

\begin{align*} \int \frac{\sqrt{x^2-9}}{x}\;dx \amp = 3 \int \sec^2(\theta)-1\;d\theta\\ \amp = 3\tan(\theta) - 3\theta + C\\ \amp = \sqrt{x^2-9} - 3\sec^{-1}\left(\frac{x}{3}\right)+C \end{align*}

🔗

🔗
From the triangle, we get $\tan(\theta) = \dfrac{x}{\sqrt{4-x^2}}\text{.}$ This gives us:

\begin{align*} \int \frac{2}{(4-x^2)^{3/2}}\;dx \amp= \frac{1}{2} \int \sec^2(\theta)\;d\theta\\ \amp = \frac{1}{2}\tan(\theta)+C\\ \amp = \frac{x}{2\sqrt{4-x^2}}+C \end{align*}

🔗

🔗
From the triangle, we get $\csc(\theta) = \dfrac{\sqrt{x^2+1}}{x}\text{.}$ This gives us:

\begin{align*} \int \frac{1}{x^2\sqrt{x^2+1}}\;dx \amp= \int \csc(\theta)\cot(\theta)\;d\theta\\ \amp = -\csc(\theta)+C\\ \amp = - \frac{\sqrt{x^2+1}}{x}+C \end{align*}

🔗

🔗

🔗

Trigonometric substitution is a pretty involved technique! Setting up the substitution is definitely not trivial. Because our substitution involves trigonometric functions, we end up with integrals of trigonometric functions that we then have to work to antidifferentiate. And substituting back to $x$ relies on us having set up a robust substitution strategy from the beginning.

🔗

It can sometimes seem like this strategy is barely relevant: the goal of it is so focussed on the specific structure of the Pythagorean identities, and these might not feel very present.

🔗

A friend of mine, though, says that once we start recognizing sums and differences of squares as being connected to Pythagoras, it’s hard to not see them.

🔗

For instance, we can go back to Theorem 7.3.3 and see the sum of squares in the denominator. Instead of doing any tricky factoring to get the $u$-substitution to work, we could try a trigonometric substitution and get the same thing!

🔗

Another friend of mine says that trigonometric substitution only exists so that we can evaluate arc length integrals (Integrals for Evaluating the Length of a Curve).

🔗

Whatever the case, this new substitution strategy should, at the very least, generalize the concept of a variable substitution in an integral to show that we can define these for a variety of purposes, all based on the kinds of structures that we’re seeing in the integrand function itself.

🔗

Practice Problems Practice Problems

1.

Explain how trigonometric substitution helps to convert sums or differences of squares to products of squares. Why is this helpful? When is it helpful?

🔗

2.

Draw a right triangle with $\sqrt{x^2-4}$ as one of the non-hypotenuse side lengths. What is the length of the hypotenuse? What about the other side length? What would be an appropriate substitution for an integral containing $\sqrt{x^2-4}\text{?}$

🔗

3.

Draw a right triangle with $\sqrt{4-x^2}$ as one of the non-hypotenuse side lengths. What is the length of the hypotenuse? What about the other side length? What would be an appropriate substitution for an integral containing $\sqrt{4-x^2}\text{?}$

🔗

4.

Draw a right triangle with $\sqrt{x^2+4}$ as the hypotenuse. What are the lengths of the other two sides? What would be an appropriate substitution for an integral containing $\sqrt{x^2+4}\text{?}$

🔗