17Calculus Integrals - Trig Substitution
Difference Between Trig Integration and Trig Substitution
Trig integration is the evaluation of integrals that already have trig functions in the integrand.
Trig substitution, covered on this page, is a technique that takes an integrand that most likely does NOT contain any trig functions, and uses some trig identities to introduce trig functions into the integrand. Once the integral is completely transformed, then trig integration is used to evaluate the integral. Once the evaluation is complete, another set of substitutions, based on the original ones, is done to convert the result back to the original variable.
This is a great technique that introduces trig into an integral which originally doesn't have any trig. This makes integration easier and sometimes possible where it wasn't before.
Recommended Books on Amazon | ||
|---|---|---|
  |
  |
  |
If you want a full lecture on trig substitution, we recommend this video from one of our favorite lecturers.
Prof Leonard - Calculus 2 Lecture 7.3: Integrals By Trigonometric Substitution [2hrs-9min-23secs]
video by Prof Leonard |
|---|
The key to using this technique is to recognize specific sets of terms in the integrand, which will tell you the associated substitution. Table one contains a summary.
| Table One - List of Factors | |
|---|---|
<\(a\) is a constant; \(x\) and \(\theta\) are variables | |
integrand | \(x^2 + a^2\) |
substitution | \(x = a\tan(\theta)\) |
identity | \( 1+\tan^2(\theta) = \sec^2(\theta)\) |
integrand | \(-x^2 + a^2\) |
substitution | \(x = a\sin(\theta)\) |
identity | \(\sin^2(\theta) + \cos^2(\theta) = 1\) |
integrand | \(x^2-a^2\) |
substitution | \(x = a\sec(\theta)\) |
identity | \( 1+\tan^2(\theta) = \sec^2(\theta)\) |
Note - - Some books and instructors may use cosecant substitution instead of secant. In this case, the substitution is \(x = a\csc(\theta)\) and the identity is \( 1+\cot^2(\theta) = \csc^2(\theta) \). | |
The terms you are looking for in the integrand may appear exactly as they are listed in table one or, more likely, they will be embedded in the integrand, like under a square root or with a power. They can appear in the numerator or the denominator of a fraction. Sometimes you may even have to complete the square in order to see the term in this form. In any case, once you have recognized the form of the term, you can choose the correct the substitution. Here are the steps to evaluate integrals using this technique.
1. Choose a substitution from table one depending on the term that appears in the integrand.
2. Draw a triangle based on the substitution.
3. Perform the substitution in the integral, simplify and evaluate.
4. Use the triangle from step 2 to convert your answer back to the original variable.
Definite Integrals
There is one possible variation to these steps. If your integral is a definite integral, you can follow the steps above and then finish the problem by evaluating the result at the limits of integration. Or you could convert your limits of integration using the substitution you chose in step 1 and then evaluate them after step 3 and skip step 4. This is up to you based on the problem statement, what you find the easiest to do and on what your instructor expects. If you are not sure, ask your instructor.
Setting Up Integrals Using Trig Substitution
Okay, before we get into the details of this technique, we need to know how to set up the integrals. Here is a great video showing how to set up each of the three types of problems.
Krista King Math - Setting Up Trigonometric Substitution [13min-12secs]
video by Krista King Math |
|---|
What To Do With Square Roots
So, what do you do when you have one of the factors in table one under a square root sign? Well, you do the same substitution as if the square root sign was not there. To understand this better, here is a video explaining this and showing an example.
Dr Chris Tisdell - Integrals With Square Root Signs [15min-33secs]
video by Dr Chris Tisdell |
|---|
Before going on, here is a quick video overview of the detail just discussed. This video may help fill in any gaps in your understanding of the basics of this technique.
Michel vanBiezen - Trig Substitution - What Is & When to Use Trig Substitution? [2min-57secs]
video by Michel vanBiezen |
|---|
Now, let's look at each substitution separately. The steps for each case are pretty much the same but we discuss each case separately to help you understand this technique. Your next logical step is to look at the tangent substitution on a separate page.
You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
|---|
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
|
Set 1 - basic identities | |||
|---|---|---|---|
\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
|
Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
|
Set 3 - double-angle formulas | |
|---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
|
Set 4 - half-angle formulas | |
|---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
To bookmark this page and practice problems, log in to your account or set up a free account.
Topics Listed Alphabetically
Single Variable Calculus |
|---|
Multi-Variable Calculus |
|---|
Differential Equations |
|---|
Precalculus |
|---|
Search Practice Problems
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
|
| |
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. |
