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#### The Fundamental Theorem of Calculus

Three Different Concepts
The Fundamental Theorem of Calculus (Part 2)
The Fundamental Theorem of Calculus (Part 1)
More FTC 1

#### The Indefinite Integral and the Net Change

Indefinite Integrals and Anti-derivatives
A Table of Common Anti-derivatives
The Net Change Theorem
The NCT and Public Policy

#### Substitution

Substitution for Indefinite Integrals
Examples to Try
Revised Table of Integrals
Substitution for Definite Integrals
Examples

#### Area Between Curves

Computation Using Integration
To Compute a Bulk Quantity
The Area Between Two Curves
Horizontal Slicing
Summary

#### Volumes

Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
More Practice

#### Integration by Parts

Integration by Parts
Examples
Integration by Parts with a definite integral
Going in Circles

#### Integrals of Trig Functions

Antiderivatives of Basic Trigonometric Functions
Product of Sines and Cosines (mixed even and odd powers or only odd powers)
Product of Sines and Cosines (only even powers)
Product of Secants and Tangents
Other Cases

#### Trig Substitutions

How Trig Substitution Works
Summary of trig substitution options
Examples
Completing the Square

#### Partial Fractions

Introduction
Linear Factors
Improper Rational Functions and Long Division
Summary

#### Strategies of Integration

Substitution
Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

#### Improper Integrals

Type 1 - Improper Integrals with Infinite Intervals of Integration
Type 2 - Improper Integrals with Discontinuous Integrands
Comparison Tests for Convergence

#### Differential Equations

Introduction
Separable Equations
Mixing and Dilution

#### Models of Growth

Exponential Growth and Decay
Logistic Growth

#### Infinite Sequences

Examples of Infinite Sequences
Limit Laws for Sequences
Theorems for and Examples of Computing Limits of Sequences
Monotonic Covergence

#### Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Test for Divergence and Other Theorems
Telescoping Sums and the FTC

#### Integral Test

The Integral Test
Estimates of Value of the Series

#### Comparison Tests

The Basic Comparison Test
The Limit Comparison Test

#### Convergence of Series with Negative Terms

Introduction, Alternating Series,and the AS Test
Absolute Convergence
Rearrangements

The Ratio Test
The Root Test
Examples

#### Strategies for testing Series

Strategy to Test Series and a Review of Tests
Examples, Part 1
Examples, Part 2

#### Power Series

Finding the Interval of Convergence
Power Series Centered at $x=a$

#### Representing Functions as Power Series

Functions as Power Series
Derivatives and Integrals of Power Series
Applications and Examples

#### Taylor and Maclaurin Series

The Formula for Taylor Series
Taylor Series for Common Functions
Adding, Multiplying, and Dividing Power Series
Miscellaneous Useful Facts

#### Applications of Taylor Polynomials

Taylor Polynomials
When Functions Are Equal to Their Taylor Series
When a Function Does Not Equal Its Taylor Series
Other Uses of Taylor Polynomials

#### Partial Derivatives

Visualizing Functions in 3 Dimensions
Definitions and Examples
An Example from DNA
Geometry of Partial Derivatives
Higher Order Derivatives
Differentials and Taylor Expansions

#### Multiple Integrals

Background
What is a Double Integral?
Volumes as Double Integrals

#### Iterated Integrals over Rectangles

How To Compute Iterated Integrals
Examples of Iterated Integrals
Cavalieri's Principle
Fubini's Theorem
Summary and an Important Example

#### Double Integrals over General Regions

Type I and Type II regions
Examples 1-4
Examples 5-7
Order of Integration

### Summary of trig substitution options

1. When $x^2 + a^2$ is embedded in the integrand, use $x = a \tan(\theta)$
(Hint:  $x^2+1$ appears in the derivative of $\tan^{-1}x$).
 Then $dx = a\, \sec^2(\theta)\, d\theta$. To convert back to $x$, use your substitution to get $\tfrac x a=\tan\theta$, and draw a right triangle with opposite side $x$, adjacent side $a$ and hypotenuse $\sqrt{x^2+a^2}$.

2. When $a^2 - x^2$ is embedded in the integrand, use $x=a \sin(\theta)$.
(Hint:  $1-x^2$ appears in the derivative of $\sin^{-1}x$.)
 Then $dx = a \cos(\theta) \,d\theta$.  To convert back to $x$, use your substitution to get $\tfrac x a=\sin(\theta)$, and draw a right triangle with opposite side $x$, hypotenuse $a$ and adjacent side $\sqrt{a^2-x^2}$.

3. When $x^2-a^2$ is embedded in the integrand, use $x=a \sec(\theta)$.
(Hint:  $x^2-1$ appears in the derivative of $\sec^{-1}x$.)
 Then $dx = a \sec(\theta) \tan(\theta)\, d\theta$. To convert back to $x$, use your substitution to get $\tfrac x a=\sec(\theta)$, and draw a right triangle with adjacent side $a$, hypotenuse $x$ and opposite side $\sqrt{x^2-a^2}$.

Examples 1 & 2:
DO:  Consider the following integrals, and determine which of the three trig substitutions is appropriate, then do the substitution.  Simplify the integrand, but do not try to evaluate it.  Don't look ahead without making an attempt.

$$\int\frac{\sqrt{9-x^2}}{x^2}\,dx,\qquad \int\frac{1}{x^2\sqrt{x^2+4}}\,dx$$

Solution 1:  The first integrand is of the form $a^2-x^2=3^2-x^2$, so we use the sine substitution.
\begin{eqnarray*}
&\int&\frac{\sqrt{9-x^2}}{x^2}\,dx \qquad\qquad\qquad {\fbox{$\, \,x\,=\,3\sin\theta,\\ dx\,=\,3\cos\theta\,d\theta$}}\\
=&\int&\frac{\sqrt{9-9\sin^2\theta}}{9\sin^2\theta}3\cos\theta\,d\theta\\\\
=&\int&\frac{3\sqrt{1-\sin^2\theta}}{9\sin^2\theta}3\cos\theta\,d\theta=\int\frac{3\sqrt{\cos^2\theta}}{9\sin^2\theta}3\cos\theta\,d\theta=\tfrac{3}{9}\cdot 3\int\frac{\cos\theta^*}{\sin^2\theta}\cos\theta\,d\theta\\\\
=&\int&\cot^2\theta\,d\theta=\int(\csc^2\theta-1)\,d\theta
\end{eqnarray*}
DO:  Finish this integration, using what we learned previously.

$^*$Notice that in general $\sqrt{\cos^2\theta}=|\cos\theta\,|$, but when using trig (inverse!) substitution, the restrictions we put on the inverse trig functions ensure that the this particular cosine value is positive, so we don't need the absolute values.  This is true for the other trig substitutions as well.

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Solution 2:  The second integrand is of the form $x^2+a^2=x^2+4$ so we use the tangent substitution.
\begin{eqnarray*}
&\int&\frac{dx}{x^2\sqrt{x^2+4}} \qquad\qquad\qquad {\fbox{$\,\, x\,=\,2\tan\theta,\\ dx\,=\,2\sec^2\theta\,d\theta$}}\\\\
=&\int&\frac{2\sec^2\theta}{4\tan^2\theta\sqrt{4\tan^2\theta+4}}\,d\theta\\\\
=&\int&\frac{2\sec^2\theta}{4\tan^2\theta\cdot 2\sqrt{\tan^2\theta+1}}\,d\theta=\frac{2}{4\cdot 2}\int\frac{\sec^2\theta}{\tan^2\theta\sqrt{\sec^2\theta}}\,d\theta\\\\
=\frac{1}{4} &\int&\frac{\sec^2\theta}{\tan^2\theta\sec\theta}\,d\theta=\frac{1}{4} \int\frac{\sec\theta}{\tan^2\theta}\,d\theta=\frac{1}{4} \int\frac{\cos\theta}{\sin^2\theta}\,d\theta
\end{eqnarray*}
DO:  Finish this integration, using what we learned previously.

DO:  Work with these two examples until you are confident of the whys, hows, and correctness of the work.