Trigonometric Substitution

Setup

Sine

$Trigonometric Substitution Example 5$

$Trigonometric Substitution Example 9$

PJ Trigonometric Substitution – Ex 3/ Part 1

PJ Trigonometric Substitution – Ex 3/ Part 2

KA Integrals: Trig Substitution 3 (long problem)

Tangent

$Trigonometric Substitution Example 3$

$Trigonometric Substitution Example 4$

$Trigonometric Substitution Example 6$

$Trigonometric Substitution Example 7$

Secant

$Trigonometric Substitution Example 1$

$Trigonometric Substitution Example 2$

$Trigonometric Substitution Example 8$

$Trigonometric Substitution Example 10$

Identifying a Trigonometric Substitution Problem

At first glance, many trigonometric substitution problems look like they might be solvable with u-substitution. However, if you try to use u-substitution, you’ll find that the integral you’re left with after the substitution is still too complicated to solve. If this is the case, you might want to consider trying trigonometric substitution.

Another obvious sign of a trigonometric substitution problem is the combination of two square terms, separated by a plus or minus sign, often within a square root sign. For example, if you have $\sqrt{81-9x^2}$ in your integral, you can see that you have two squares, $81$ is the square of $9$ and $9x^2$ is the square of $3x$ . These two terms are separated by a minus sign, and they’re within a square root sign. Again, the square root is not necessary, but this is a great indication that you may be dealing with a trigonometric substitution problem.

Solving a Trigonometric Substitution Problem

To solve an integral using trigonometric substitution, you should perform the following steps, each of which we’ll cover in more detail later on in this section:

Identification and setup
Substitution and simplification
Integration
Back substitution

1. Identification and Setup

The first thing you need to do is identify the kind of trigonometric substitution you’re dealing with. It will either be a substitution involving sine, tangent, or secant, and your job is to figure out which one it is. Luckily, identifying which one to use is very straightforward. Remember before how we talked about those two square terms that are separated by a plus or minus sign? Those are the two terms you’ll want to focus on. One should be a constant, the $81$ from above, and one should involve the variable, the $9x^2$ from above.

Is there a plus sign between them? Then you’ll use the tangent substitution. If the sign between the two terms is a negative sign and the constant is first (constant term minus the variable term), then you’ll use the sine substitution. If the sign between them is negative and the variable term is first (variable term minus the constant term), then you’ll use the secant substitution. Now that you’ve identified which one of the substitutions you’ll use, refer to the chart on this page as a quick reference guide for the setup of your trigonometric substitution problem.

Setup is one of the most important parts of solving a trigonometric substitution problem. You never know exactly which pieces of the setup steps you’ll need later on in the problem, and it will help you a lot to work through these quickly if you get in the habit of calculating each piece of the setup process. Then, when you need to reference the information later on, you’ll have in ready. There are six pieces to solve for, plus you’ll need to draw a triangle. So there are seven total pieces that you’ll need.

The first thing to do is to identify $a$ and $u$ . To find $a$ , take the square root of the constant term. To find $u$ , take the square root of the variable term. So, if we’re referencing our example from earlier where the integral included $\sqrt{81-9x^2}$ , the constant term is $81$ and the variable term is $9x^2$ . To find $a$ , take the square root of $81$ , and you’ll calculate that $a=9$ . To find $u$ , take the square root of $9x^2$ , and you’ll calculate that $u=3x$ .

If you’re making a sine substitution, then you’ll substitute $u=a\sin{\theta}$ ( $3x=9\sin{\theta}$ which is equal to $x=3\sin{\theta}$ ).
If you’re making a tangent substitution, then you’ll substitute $u=a\tan{\theta}$ .
If you’re making a secant substitution, then you’ll substitute $u=a\sec{\theta}$ .

Now that you’ve identified the substitution you’ll make, you need to pick the identity you’ll use in conjunction with your substitution.

If you’re making a sine substitution, then you’ll use the identity $1-\sin^2{\theta}=\cos^2{\theta}$ .
If you’re making a tangent substitution, then you’ll use the identity $1+\tan^2{\theta}=\sec^2{\theta}$ .
If you’re making a secant substitution, then you’ll use the identity $\sec^2{\theta}-1=\tan^2{\theta}$ .

You’re already through the first three steps, and you just have three more calculations to make to finish your setup. From the substitution you calculated earlier, solve for the trigonometric identity.

If your substitution was $u=a\sin{\theta}$ ( $x=3\sin{theta}$ ), then solve for $\sin{\theta}$ by dividing both sides by $a$ . You’ll get $\sin{\theta}=\frac{u}{a}$ , or in the case of our example, $\sin{\theta}=\frac{x}{3}$ .
If your substitution was $u=a\tan{\theta}$ , then solve for $\tan{\theta}$ by dividing both sides by $a$ . You’ll get $\tan{\theta}=\frac{u}{a}$ .
If your substitution was $u=a\sec{\theta}$ , then solve for $\sec{\theta}$ by dividing both sides by $a$ . You’ll get $\sec{\theta}=\frac{u}{a}$ .

For the next setup step, find the derivative of the substitution equation you found earlier.

If your substitution was $u=a\sin{\theta}$ ( $x=3\sin{\theta}$ ), then the derivative will be $du=a\cos{\theta}\ d\theta$ , or in the case of our example, $dx=3\cos{\theta}$ .
If your substitution was $u=a\tan{\theta}$ , then the derivative will be $du=a\sec^2{\theta}\ d\theta$ .
If your substitution was $u=a\sec{\theta}$ , then the derivative will be $du=a\sec{\theta}\tan{\theta}\ d\theta$ .

The final setup step for your trigonometric substitution problem is to solve your substitution equation for $\theta$ .

If your substitution was $u=a\sin{\theta}$ ( $x=3\sin{\theta}$ ), then solving for $\theta$ will give you $\theta=\sin^{-1}\frac{u}{a}$ .
If your substitution was $u=a\tan{\theta}$ , then solving for $\theta$ will give you $\theta=\tan^{-1}\frac{u}{a}$ .
If your substitution was $u=a\sec{\theta}$ , then solving for $\theta$ will give you $\theta=\sec^{-1}\frac{u}{a}$ .

Now that you finally have all six pieces of the trigonometric substitution in place, your final setup step is to draw your triangle. Depending on whether you’re making a sine, tangent, or secant substitution, you’ll label different sides of your triangle with $a$ , $u$ , and the part of the integral that you identified in the first place for trigonometric substitution. See the table on this page for a more detailed look at how to label the different sides of your triangle.

2. Substitution and simplification

Once you complete the setup portion of your trigonometric substitution problem, you need to start making substitutions into your integral. During setup, you set $x=a\sin{\theta}$ or $x=a\tan{\theta}$ or $x=a\sec{\theta}$ , and you calculated the derivative of $x$ to get $dx$ . Now you need to substitute for $x$ and $dx$ in your integral.

After making the substitutions, simplify as much as you can. Part of simplification will be to use the identity that you wrote down during setup. Remember,

If you’re making a sine substitution, then you’ll use the identity $1-\sin^2{\theta}=\cos^2{\theta}$ .
If you’re making a tangent substitution, then you’ll use the identity $1+\tan^2{\theta}=\sec^2{\theta}$ .
If you’re making a secant substitution, then you’ll use the identity $\sec^2{\theta}-1=\tan^2{\theta}$ .

If you made the substitution of $x$ and $dx$ correctly, you should end up with a piece of one of the identities above. Substitute for the identity, and then continue simplifying the integral as much as possible.

3. Integration

After simplifying as much as possible, it’s time to actually integrate the function. Go ahead and evaluate the integral. Don’t forget to add $C$ to account for the constant of integration.

4. Back substitution

Once the integration is complete, you’ll have a function in terms of $\theta$ . You need to back-substitute for $\theta$ to get the function in terms of $x$ . You may have to substitute for $\theta$ , or you may have to use your triangle to make substitutions, depending on how your function looks after integration.