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

Area Between Curves

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


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

Integration by Parts

Integration by Parts
Integration by Parts with a definite integral
Going in Circles
Tricks of the Trade

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
Completing the Square

Partial Fractions

Linear Factors
Irreducible Quadratic Factors
Improper Rational Functions and Long Division

Strategies of Integration

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

Separable Equations
Mixing and Dilution

Models of Growth

Exponential Growth and Decay
Logistic Growth

Infinite Sequences

Approximate Versus Exact Answers
Examples of Infinite Sequences
Limit Laws for Sequences
Theorems for and Examples of Computing Limits of Sequences
Monotonic Covergence

Infinite Series

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

Integral Test

Road Map
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

The Ratio and Root Tests

The Ratio Test
The Root Test

Strategies for testing Series

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

Power Series

Radius and Interval of Convergence
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

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


In this video, we work three examples, one with $x=a \tan(\theta)$, one with $x = a\sin(\theta)$, and one with $x = a \sec(\theta)$.  Worked out solutions are written below the video.

Examples from the video.

DO:  After watching the video, write down and work these examples on your own, slowly, thinking of the whys and hows of each step.

Example 1:  $\int \bigl(4+x^2\bigr)^{-3/2}\, dx$
DO:  Work through before looking ahead.

Solution 1:  $\displaystyle\int \bigl(4+x^2\bigr)^{-3/2}\, dx \overset{\fbox{$ \,\, x\,=\,2\tan\theta\\dx\,=\,2\sec^2\theta\,d\theta$}}{=}\int \bigl(4 + 4\tan^2\theta\bigr)^{-3/2} 2 \sec^2\theta \,d\theta=\int \bigl(4 \sec^2\theta\bigr)^{-3/2} 2 \sec^2(\theta) \,d\theta$

$\displaystyle =4^{-3/2}\cdot 2\int\sec^{-3}\theta\sec^2\theta\,d\theta=\frac{2}{8}\int\sec^{-1}\,d\theta=\frac{1}{4}\int \frac{d\theta}{\sec\theta} = \frac{1}{4}\int \cos\theta\, d\theta = \frac{1}{4}\sin\theta+C.$
Consider our answer above.  In order to convert back into terms of $x$, we must figure out what $\sin(\theta)$ is in terms of $x$. By rewriting our original substitution we see that $\tfrac x 2=\tan\theta$.  Use this to draw a right triangle, with opposite side $x$ and adjacent side $a=2$.  The hypotenuse is then $\sqrt{a^2+x^2}=\sqrt{4+x^2}$.  We need to find $\sin\theta$ in terms of $x$, and we see from the triangle that $\sin\theta=\frac{x}{\sqrt{x^2+4}}$.
So $\displaystyle\int\left(4+x^2\right)^{-3/2}\,dx=\frac{1}{4}\sin\theta+C=\frac{x}{4\sqrt{x^2+4}}+C$.

Example 2:  $\int \sqrt{9-x^2}\, dx$
DO:  Work through before looking ahead.

Solution 2: $\displaystyle\int \sqrt{9-x^2}\, dx\overset{ \fbox{$ \,\,x\,=\,3 \sin\theta\\dx\,=\,3\cos\theta\,d\theta$} }{=} \int \left(\sqrt{9-9\sin^2\theta}\right)3\cos\theta\,d\theta=3\cdot 3\int\sqrt{\cos^2\theta}\cos\theta \,d\theta=9\int\cos^2\theta \,d\theta$

$\displaystyle\quad =9\int\frac{1+\cos(2\theta)}{2}\,d\theta=\frac{9}{2}\int(1+\cos(2\theta))\,d\theta =\frac{9}{2} \left(\theta+\frac{\sin(2\theta)}{2}\right)+C=\frac{9}{2}\bigl(\theta+\sin\theta\cos\theta\bigr)+C $$
Here we have used the methods of the last learning module to evaluate the trig integral, including the handy trig identities for $\cos^2\theta$ and $\sin(2\theta)$.  (You need to know these by heart).   We look at the terms in our final answer above.  We use the triangle to convert $\sin\theta\cos\theta$ back into terms of $x$.  Finally, we must write $\theta$ in terms of $x$.  We use our original substitution: $\frac{x}{3}=\sin x$ gives us $\sin^{-1}(\tfrac x 3)=\theta$.
So we have $\displaystyle\int \sqrt{9-x^2}\, dx=\frac{9}{2}\left( \sin^{-1} \left (\frac{x}{3} \right ) + \frac{x}{3}\frac{\sqrt{9-x^2}}{3}\right)+ C=\frac{9}{2} \sin^{-1} \left (\frac{x}{3} \right ) + \frac{x \,\sqrt{9-x^2}}{2} + C$

Example 3: $\displaystyle\int \frac{dx}{\sqrt{4x^2-1}}$
DO:  Work through before looking ahead.

Solution 3:  $\displaystyle\int \frac{dx}{\sqrt{4x^2-1}} \overset{ \fbox{$ 2x\,=\, \sec\theta\\2dx\,=\,\sec\theta\tan\theta\,d\theta$} }{=} \int \frac{\frac{1}{2}\sec\theta\tan\theta\,d\theta}{\sqrt{\sec^2\theta-1}}
\overset{ \fbox{$ \sec^2\theta-1\,=\,\tan^2\theta$} }{=} \frac{1}{2}\int \frac{\sec\theta\tan\theta\,d\theta}{\sqrt{\tan^2\theta}}$

$\displaystyle= \frac{1}{2}\int \frac{\sec\theta\tan\theta\,d\theta}{\tan\theta}=\frac{1}{2}\int\sec\theta\,d\theta=\frac{1}{2} \ln\bigl\lvert\sec\theta+\tan\theta\bigr\rvert +C. $

We now convert to terms of $x$:  $\sec\theta=2x$, so $\cos\theta=\frac{1}{2x}$.  The triangle could therefore have adjacent side of 1, and hypotenuse of $2x$, and the opposite side $\sqrt{4x^2-1}$.  We can also have adjacent of $\frac{1}{2}$ and hypotenuse of $x$. (Why?)  This gives the opposite side the value of $\sqrt{x^2-\frac{1}{4}}$, as in the diagram.  Either way is fine.  Looking at our values above, we only need deal with $\tan \theta$, since we know $\sec\theta=2x$.  From the triangle, $\tan\theta=\frac{\sqrt{x^2-1/4}}{1/2}$.  If we had used the other triangle, we would get $\tan\theta=\frac{\sqrt{4x^2-1}}{1}$ -- are these the same values?

So $\displaystyle\int \frac{dx}{\sqrt{4x^2-1}} =\frac{1}{2} \ln \bigl\lvert \,2x + \sqrt{4x^2-1}\,\bigr\rvert + C.$