Limits Derivatives Integrals Infinite Series Parametrics Polar Coordinates Conics
Limits
Epsilon-Delta Definition
Finite Limits
One-Sided Limits
Infinite Limits
Trig Limits
Pinching Theorem
Indeterminate Forms
L'Hopitals Rule
Limits That Do Not Exist
Continuity & Discontinuities
Intermediate Value Theorem
Derivatives
Power Rule
Product Rule
Quotient Rule
Chain Rule
Trig and Inverse Trig
Implicit Differentiation
Exponentials & Logarithms
Logarithmic Differentiation
Hyperbolic Functions
Higher Order Derivatives
Differentials
Slope, Tangent, Normal...
Linear Motion
Mean Value Theorem
Graphing
1st Deriv, Critical Points
2nd Deriv, Inflection Points
Related Rates Basics
Related Rates Areas
Related Rates Distances
Related Rates Volumes
Optimization
Integrals
Definite Integrals
Integration by Substitution
Integration By Parts
Partial Fractions
Improper Integrals
Basic Trig Integration
Sine/Cosine Integration
Secant/Tangent Integration
Trig Integration Practice
Trig Substitution
Linear Motion
Area Under/Between Curves
Volume of Revolution
Arc Length
Surface Area
Work
Moments, Center of Mass
Exponential Growth/Decay
Laplace Transforms
Describing Plane Regions
Infinite Series
Divergence (nth-Term) Test
p-Series
Geometric Series
Alternating Series
Telescoping Series
Ratio Test
Limit Comparison Test
Direct Comparison Test
Integral Test
Root Test
Absolute Convergence
Conditional Convergence
Power Series
Taylor/Maclaurin Series
Interval of Convergence
Remainder & Error Bounds
Fourier Series
Study Techniques
Choosing A Test
Sequences
Infinite Series Table
Practice Problems
Exam Preparation
Exam List
Parametrics
Parametric Curves
Parametric Surfaces
Slope & Tangent Lines
Area
Arc Length
Surface Area
Volume
Polar Coordinates
Converting
Slope & Tangent Lines
Area
Arc Length
Surface Area
Conics
Parabolas
Ellipses
Hyperbolas
Conics in Polar Form
Vectors Vector Functions Partial Derivatives/Integrals Vector Fields Laplace Transforms Tools
Vectors
Unit Vectors
Dot Product
Cross Product
Lines In 3-Space
Planes In 3-Space
Lines & Planes Applications
Angle Between Vectors
Direction Cosines/Angles
Vector Projections
Work
Triple Scalar Product
Triple Vector Product
Vector Functions
Projectile Motion
Unit Tangent Vector
Principal Unit Normal Vector
Acceleration Vector
Arc Length
Arc Length Parameter
Curvature
Vector Functions Equations
MVC Practice Exam A1
Partial Derivatives
Directional Derivatives
Lagrange Multipliers
Tangent Plane
MVC Practice Exam A2
Partial Integrals
Describing Plane Regions
Double Integrals-Rectangular
Double Integrals-Applications
Double Integrals-Polar
Triple Integrals-Rectangular
Triple Integrals-Cylindrical
Triple Integrals-Spherical
MVC Practice Exam A3
Vector Fields
Curl
Divergence
Conservative Vector Fields
Potential Functions
Parametric Curves
Line Integrals
Green's Theorem
Parametric Surfaces
Surface Integrals
Stokes' Theorem
Divergence Theorem
MVC Practice Exam A4
Laplace Transforms
Unit Step Function
Unit Impulse Function
Square Wave
Shifting Theorems
Solve Initial Value Problems
Prepare For Calculus 1
Trig Formulas
Describing Plane Regions
Parametric Curves
Linear Algebra Review
Word Problems
Mathematical Logic
Calculus Notation
Simplifying
Practice Exams
More Math Help
Tutoring
Tools and Resources
Learning/Study Techniques
Math/Science Learning
Memorize To Learn
Music and Learning
Note-Taking
Motivation
Instructor or Coach?
Books
Math Books

You CAN Ace Calculus

17calculus > integrals > trig substitution

### Calculus Main Topics

Integrals

Integral Applications

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

Integration Technique - Trigonometric 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.

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

integrand

substitution

identity

tangent

$$x^2 + a^2$$

$$x = a\tan(\theta)$$

$$1+\tan^2(\theta) = \sec^2(\theta)$$

sine

$$-x^2 + a^2$$

$$x = a\sin(\theta)$$

$$\sin^2(\theta) + \cos^2(\theta) = 1$$

secant

$$x^2-a^2$$

$$x = a\sec(\theta)$$

$$1+\tan^2(\theta) = \sec^2(\theta)$$

a is a constant; x and θ are variables

table one - list of factors

a is a constant; x and θ are variables

tangent

integrand

$$x^2 + a^2$$

substitution

$$x = a\tan(\theta)$$

identity

$$1+\tan^2(\theta) = \sec^2(\theta)$$

sine

integrand

$$-x^2 + a^2$$

substitution

$$x = a\sin(\theta)$$

identity

$$\sin^2(\theta) + \cos^2(\theta) = 1$$

secant

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 you use 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.

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

The Substitutions

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.

### Tangent Substitution     x2 + a2   →   x = a tan(θ)

Overview
For this substitution, we also use the equation $$1+\tan^2(\theta) = \sec^2(\theta)$$. Notice that when we perform the substitution, we have

$$\displaystyle{ \begin{array}{rcl} a^2+x^2 & = & a^2 + (a\tan(\theta))^2 \\ & = & a^2 + a^2 \tan^2(\theta) \\ & = & a^2 (1 + \tan^2(\theta)) \\ & = & a^2 \sec^2(\theta) \\ \end{array} }$$

In the last equation we end up with two squared terms, which we can easily take the square root of, for example, to simplify.

Triangle
As mentioned in step 2 above, we need to build a triangle based on our choice of substitution, in this case $$x = a\tan(\theta) \to \tan(\theta) = x/a$$. So, we draw a right triangle, select one of the other two angles to be $$\theta$$ and label $$a$$ and $$x$$ accordingingly. Then we use the Pythagorean Theorem to get the expression for the other side. This gives us the triangle on the right.

Next, we substitute in the integral for $$x = a\tan(\theta))$$ and $$dx = a\sec^2(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Practice
Click here to go to a list of problems to practice this technique.

### Sine Substitution     – x2 + a2  →  x = a sin(θ)

Overview
The other equation we use in this substitution is $$\sin^2(\theta) + \cos^2(\theta) = 1$$. Using the substitution $$x = a\sin(\theta)$$ in $$a^2-x^2$$ gives us

$$\displaystyle{ \begin{array}{rcl} a^2-x^2 & = & a^2 - (a\sin(\theta))^2 \\ & = & a^2 - a^2 \sin^2(\theta) \\ & = & a^2(1-\sin^2(\theta)) \\ & = & a^2 \cos^2(\theta) \end{array} }$$

This may not seem like much but think about this. If the term $$a^2-x^2$$ is under a square root, then the last term above could be simplified quite a bit.

$$\sqrt{a^2-x^2 } = \sqrt{a^2 \cos^2(\theta)} = a\cos(\theta)$$

Notice we lost the square root and now have a term that, depending on it's location in the integrand, is much easier to integrate.

Triangle
As mentioned in the steps above, we need to draw a triangle based on our choice of substitution, in this case $$x = a\sin(\theta) \to \sin(\theta) = x/a$$. Using this last equation, we draw a right triangle and label the angle $$\theta$$, then sides $$x$$ and $$a$$. Then we use the Pythagorean Theorem to get the expression for the third side.

Next, we substitute in the integral for $$x = a\sin(\theta)$$ and $$dx = a\cos(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Practice
Click here to go to a list of problems to practice this technique.

### Secant Substitution     x2 – a2    →   x = a sec(θ)

Overview
In this case, we have two choices, to use secant or cosecant. Fortunately, both should work. So you may choose either based on what you are comfortable with, what your instructor requires and any other restraining factors. We will discuss the use of cosecant but the use of secant parallels this and we think you can easily extrapolate and alter the discussion to fit secant.

Similar to the other two cases above, we have the equation $$1+\cot^2(\theta) = \csc^2(\theta)$$ that we will use to simplify. Performing the substitution, we have

$$\displaystyle{ \begin{array}{rcl} -a^2+x^2 & = & -a^2 + (a\csc(\theta))^2 \\ & = & -a^2 + a^2 \csc^2(\theta) \\ & = & a^2(-1+\csc^2(\theta)) \\ & = & a^2 \cot^2(\theta) \end{array} }$$

This is a much simpler expression, especially if the $$-a^2+x^2$$ is under a square root.

Triangle
We use $$x = a~\csc(\theta) \to \csc(\theta) = x/a \to 1/\sin(\theta) = x/a \to \sin(\theta) = a/x$$   to build the triangle on the right. We choose one angle to label $$\theta$$, label the other two sides using the last equation and then calculate the expression for the third side using the Pythagorean Theorem.

Next, we substitute in the integral for $$x = a\csc(\theta)$$ and $$dx = a\csc(\theta)\cot(\theta)~d\theta$$. This last substitution is important to remember since we can't just replace $$dx$$ with $$d\theta$$ and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into $$x$$. It is not correct to leave $$\theta$$'s in the final answer.

Practice
Click here to go to a list of problems to practice this technique.

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]

Setting Up Integrals Using Trig Substitution

Okay, now that you know what substitution to use, let's learn how to use those substitutions. Here is a great video showing how to set up each of the three types of problems discussed above.

 Krista King Math - Setting Up Trigonometric Substitution

Expanding The Use of This Technique

In the discussion so far on this page, we have assumed that the $$x$$ variable term is always exactly $$x$$. There is a generalization of this where you might have $$[f(x)]^2 + a^2$$ and $$f(x) \neq x$$. This technique still works and you just change the formulas as follows.

table two

integrand

substitution

tangent

$$[f(x)]^2+a^2$$

$$f(x) = a\tan(\theta)$$

sine

$$-[f(x)]^2+a^2$$

$$f(x) = a\sin(\theta)$$

secant

$$[f(x)]^2-a^2$$

$$f(x) = a\sec(\theta)$$

f(x) can be pretty much anything (but not a constant, so check that it contains at least one x). Some examples are

table three

integrand

substitution

$$e^{2x}+9$$

$$e^x = 3\tan(\theta)$$

$$-(x+2)^2+16$$

$$x+2 = 4\sin(\theta)$$

$$4x^2-25$$

$$2x = 5\sec(\theta)$$

### Search 17Calculus

practice filters

tangent substitution

sine substitution

secant/cosecant substitution

Practice Problems

Instructions - - Unless otherwise instructed, evaluate the following integrals. Give all answers in exact, simplified form.

 Level A - Basic

Practice A01

$$\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } }$$

solution

Practice A02

$$\displaystyle{ \int{ \frac{dx}{\sqrt{81-9x^2}} } }$$

solution

Practice A03

$$\displaystyle{ \int_{0}^{10}{\frac{dx}{x^2+25} } }$$

solution

Practice A04

$$\displaystyle{ \int_{10\sqrt{3}}^{10}{\frac{1}{\sqrt{x^2-64}}~dx } }$$

solution

Practice A05

$$\displaystyle{\int{\frac{dx}{\sqrt{9x^2+4}}}}$$

solution

Practice A06

$$\displaystyle{ \int_{0}^{2}{\frac{dx}{\sqrt{x^2+4}}}}$$

solution

Practice A07

$$\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }$$

solution

Practice A08

$$\displaystyle{\int{\frac{dx}{(4x^2+9)^2}}}$$

solution

Practice A09

$$\displaystyle{ \int{ \frac{dx}{(a^2+x^2)^{3/2}}} }$$

solution

Practice A10

$$\displaystyle{ \int_{0}^{3}{\sqrt{9-x^2}~dx} }$$

solution

 Level B - Intermediate

Practice B01

Solve the initial value problem $$\displaystyle{\frac{dy}{dt}=\frac{1}{25+9t^2},~~~y(5/3)=\frac{\pi}{30}}$$

solution

Practice B02

$$\displaystyle{\int{\frac{x^3}{\sqrt{16-x^2}}~dx}}$$

solution

Practice B03

$$\displaystyle{\int_{\ln(3/2)}^{\ln 3}{e^x\sqrt{9-e^{2x}}~dx}}$$

solution

Practice B04

$$\displaystyle{\int{\frac{x^2}{\sqrt{25-x^2}}~dx}}$$

solution

Practice B05

$$\displaystyle{\int{\frac{dx}{(4x^2-1)^{3/2}}}}$$

solution

Practice B06

$$\displaystyle{\int_{4}^{8}{\frac{dx}{x^2\sqrt{x^2-16}}}}$$

solution

Practice B07

$$\displaystyle{\int{\frac{\sqrt{x^2-25}}{x}~dx}}$$

solution

Practice B08

$$\displaystyle{\int{\sqrt{x^2+4}~dx}}$$

solution

Practice B09

$$\displaystyle{ \int{ \frac{dx}{x^2+4x+1} } }$$

solution

Practice C01

$$\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }$$

solution

Practice C02

$$\displaystyle{ \int_{0}^{4}{ x^2 \sqrt{16-x^2} ~dx } }$$

solution

Practice C03

$$\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }$$

solution

Practice C04

$$\displaystyle{ \int{\sqrt{9+16x^2}~dx} }$$

solution

Practice C05

$$\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }$$

solution

Practice C06

$$\displaystyle{ \int{ \frac{dx}{(5-4x-x^2)^{5/2}} } }$$

$$\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }$$