Integration Example Problem 1

The following indefinite integral is a great exercise for students in introductory calculus since it requires most of the techniques they learn in order to integrate.

First rewrite the integral in order to use a trigonometric substitution.

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \int \sqrt{ \left( x + 3 \right)^2 – 4 } \ dx $$

Substituting and simplifying yields

$$ x = 2 \sec \theta – 3 \hspace{ 20 mm } dx = 2 \sec \theta \tan \theta \ d \theta $$
$$ 2 \int \sqrt{ 4 \sec^2 \theta – 4 } \sec \theta \tan \ \theta d \theta $$
$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 4 \left( \int \left( \sec^2 \theta – 1 \right) \sec \theta \ d \theta \right) $$

From here we need to use integration by parts and the antiderivative of \( \sec \theta \). I assume the \( \int \sec \theta \ d \theta \) is known to the reader. If not, click here for a pop-out window solution.

$$4 \left( \int \sec^3 \theta \ d \theta – \int \sec \theta \ d \theta \right) = 4 \left( \int \sec^3 \theta \ d \theta – \ln \left| \sec \theta + \tan \theta \right| + C \right) $$

Using integration by parts

$$ \int \sec^3 \theta \ d \theta = \int \sec^2 \theta \sec \theta \ d \theta $$
$$ u = \sec \theta \hspace{20 mm } du = \sec \theta \tan \theta \ d \theta $$
$$ dv = \sec^2 \theta \ d \theta \hspace{20 mm } v = \tan \theta $$
$$ \int \sec^3 \theta \ d \theta = \sec \theta \tan \theta \ – \int \tan^2 \theta \sec \theta \ d \theta $$
$$ \int \sec^3 \theta \ d \theta = \sec \theta \tan \theta \ – \int \sec^3 \theta + \int \sec \theta \ d \theta $$
$$ \int \sec^3 \theta \ d \theta = \frac{1}2 \sec \theta \tan \theta + \frac{1}2 \ln \left| \sec \theta + \tan \theta \right| $$

Remembering to multiply through by 4

$$ 4 \int \sec^3 \theta \ d \theta = 2 \sec \theta \tan \theta + 2 \ln \left| \sec \theta + \tan \theta \right| $$

Reassembling the pieces of the original integral and simplifying

$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 2 \sec \theta \tan \theta + 2 \ln \left| \sec \theta + \tan \theta \right| \ – 4 \ln \left| \sec \theta + \tan \theta \right| + C $$
$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 2 \sec \theta \tan \theta – 2 \ln \left| \sec \theta + \tan \theta \right| + C $$

Substituting back in for \( x \)

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| \frac{ x + 3 }{2} + \frac{ \sqrt{ x^2 + 6x + 5 } }2 \right| + K $$

Using log rules the ratio can be removed from the argument of the log function.

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \left( \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| – \ln \left| 2 \right| \right) + K $$
$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| + 2 \ln \left| 2 \right| + K $$
$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| + C_1$$

where \( C_1 = 2 \ln \left| 2 \right| + K \)

It’s Official: I’m Giving Up on MS Word in Favor of LaTeX

All my future mathematical writing will be completed using LaTeX (pronounced lay-tek). LaTeX is a high level markup language based on the low level computer language TeX (pronounced Tek) invented by Donald Knuth. LaTeX is the professional standard for typesetting mathematical and technical documents.

MS Word is great software for the nontechnical writers, and I may use MS Word from time to time for any writing that I do that does not contain mathematical notation. However, I have decided not to waste anymore time trying to make Word documents that contain equations; moreover, I am going to begin teaching it to my calculus II students so that they can complete their test corrections in a professional typed format. I will be encouraging my other students to begin learning it.

LaTeX takes persistence to learn in the beginning; however, if you do not give up you’ll be rewarded with professionally typeset documents. There are a plethora of online TeX communities and free “how-to” documents on the world wide web. I am still learning myself. Visit my Downloads page to download some sample documents that I have put together.

If you would like to get started you will need to download and install a TeX compiler and watch a few youtube videos. Visit the TeX Users Group or the Comprehensive TeX Archive for more information.

Limit: Geometric Sequence

If \( a_n = r^n \) is an infinite geometric sequence of real numbers and \( 0 < r < 1 \) then

$$ \lim_{ n \rightarrow \infty } r^n = 0 $$

Without loss of generality \( r \) can be defined as \( r = \frac{1}{ 1 + c } \) for \( c \gt 0 \).

Using Bernoulli’s Inequality

$$ r^n = {\left( \frac{1}{1 + c} \right)}^n = \frac{1}{\left( 1 + c \right)^n} \le \frac{1}{1 + nc} \le \frac{1}{nc} $$

Reduction Formulas

\( \int \! \sin^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \sin^n x \ dx = \int \! \left( \sin^{n-1} x \right) \sin x \ dx \)

\( u = \sin^{n-1} x \)

\( du = \left(n-1 \right) \sin^{n-2} x \cos x \ dx \)

\( dv = \sin x \ dx \)

\( v = – \cos x \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \cos^2 x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \left( 1 – \sin^2 x \right) \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x – \sin^{\left(n-2\right)+2} x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x – \sin^n x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx – \left(n-1 \right) \int \! \sin^n x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx + \left(n-1 \right) \int \! \sin^n x \ dx= -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \)

\( n \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = \frac{1}n \left( -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \right) \)

 

\( \int \! \cos^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \cos^n x \ dx = \int \! \left( \cos^{n-1} x \right) \cos x \ dx \)

\( u = \cos^{n-1} x \)

\( du = – \left(n-1 \right) \cos^{n-2} x \sin x \ dx \)

\( dv = \cos x \ dx \)

\( v = \sin x \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \sin^2 x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \left( 1 – \cos^2 x \right) \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x – \cos^{\left(n-2\right)+2} x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x – \cos^n x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx – \left(n-1 \right) \int \! \cos^n x \ dx \)

\(n \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \frac{1}n \left( \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx \right) \)

 

\( \int \! \tan^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 2 \)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \tan^2 x \ dx\)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \left( \sec^2 x – 1 \right) dx\)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \sec^2 x \ dx – \int \! \tan^{n-2}x \ dx \)

\( \int \! \tan^{n-2}x \sec^2 x \ dx\)

\( u = \tan x \)

\( du = \sec^2 x \ dx\)

\( \int \! u^{n-2} \ du = \frac{u^{n-1}}{n-1} = \frac{ \tan^{n-1}x}{n-1} \)

\( \int \! \tan^n x \ dx = \frac{ \tan^{n-1}x}{n-1} – \int \! \tan^{n-2}x \ dx \)

 

\( \int \! \sec^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \sec^n x \ dx = \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx \)

\( u = \sec^{n-2} x \)

\( du = \left(n-2 \right) \sec^{n-3} x \sec x \tan x \ dx \)

\( dv = \sec^2 x \ dx \)

\( v = \tan x \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \tan^2 x \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \left( \sec^2 x – 1 \right) \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \sec^n x – \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \sec^n x \ dx + \left(n-2 \right) \int \! \sec^{n-2} x \ dx \)

\( \left(n-1 \right) \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x + \left( n-2 \right) \int \! \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \frac{1}{n-1} \left( \tan x \sec^{n-2} x + \left( n-2 \right) \int \! \sec^{n-2} x \ dx \right) \)

 

\( \int \! \csc^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \csc^n x \ dx = \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx \)

\( u = \csc^{n-2} x \)

\( du = \left(n-2 \right) \csc^{n-3} x \left(-\csc x \cot x \right) \ dx \)

\( dv = \csc^2 x \ dx \)

\( v = -\cot x \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \cot^2 x \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \left( \csc^2 x – 1 \right) \sec^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \csc^n x – \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \csc^n x \ dx + \left(n-2 \right) \int \! \csc^{n-2} x \ dx \)

\( \left(n-1 \right) \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x + \left( n-2 \right) \int \! \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = \frac{1}{n-1} \left( -\cot x \csc^{n-2} x + \left( n-2 \right) \int \! \csc^{n-2} x \ dx \right) \)

L’Hopital’s Rule

$$ \lim_{ x \rightarrow 0^{+} } x^{ \sin x } $$

$$ y = x^{ \sin x } $$

$$ \ln y = ln x^{ \sin x } $$

$$ \ln y = \sin x \ln x $$

$$ \lim_{ x \rightarrow 0^{+} } \ln y = \lim_{x \rightarrow 0^{+}} \frac{ \ln x }{ \frac{ 1 }{ \sin x } } $$

$$ \lim_{ x \rightarrow 0^{+} } \ln y = \lim_{x \rightarrow 0^{+}} \frac{ \ln x }{ \frac{ 1 }{ \sin x } } = lim_{x \rightarrow 0^{+}} \frac{ \frac{ 1 }{ x }}{ \frac{ – cos x }{ sin^2 x }} $$

$$ \lim_{ x \rightarrow 0^{+} } \ln y = \lim_{x \rightarrow 0^{+}} \frac{ – \sin^2 x }{ x \cos x } = \lim_{x \rightarrow 0^{+}} \frac{ – 2 \sin x \cos x }{ \cos x – x \sin x } = 0 $$

$$ e^{ \lim_{ x \rightarrow 0^{+} } \ln y } = e^{ 0 } $$

$$ \lim_{ x \rightarrow 0^{ + }} e^{ \ln y } = 1 $$

$$ \lim_{ x \rightarrow 0^{+} } x^{ \sin x } = 1 $$

Fun With the Field Axioms

A1:  a + b = b + a

A2:  left( a + b right) + c = a + left( b + c right)

A3:  a + 0 = a

A4: a + left( -a right) = 0

 

M1:  ab = ba

M2:  a left( bc right) = left( ab right) c

M3:  a cdot 1 = a

M4:  a cdot a^{-1} = 1

D1:  a left( b+ c right)= ab + ac

 

Theorems Proved on the next page:

  1. 0 cdot a = 0
  2. If a + b = a then b = 0
  3. – left( -a right) = a
  4. If a + b = 0 then a = -b
  5. – left( a + b right) = left( -a right) + left( -b right)
  6. left( -a right) left( -b right) = ab

Geometric Series Formulas

Let us consider the geometric series \( \sum_{k=t}^m ar^{k} \). The first \( n \) partial sums are

\(S_t = a r^t\)
\(S_{ t + 1 } = a r^t + a r^{ t + 1 }\)
\(S_{ t + 2 } = a r^t + a r^{ t + 1 } + a r^{ t + 2 }\)
\( \vdots \)
\(S_n = a r^t + a r^{ t + 1 } + a r^{ t + 2 } + \cdots + a r^{ t + j }\)

A formula for \( S_n\) by using the following cleverness; Factor \( r^t \) from the nth partial sum yielding

$$S_n = r^t \left( a + a r^t + a r^{ t + 1 } + \cdots + a r^{ j } \right)$$

which is equal to

$$S_n = r^t \left( \sum_{k=0}^{ j } ar^{k} \right) = \frac{ r^t \left( a – a r^{ j + 1} \right)}{ 1 – r }$$

Now consider the geometric series \( \sum_{k=t}^m ar^{k – 1} \). The first \( n \) partial sums are

\(S_t = a r^{ t – 1 }\)
\(S_{ t + 1 } = a r^{ t – 1 } + a r^t \)
\(S_{ t + 2 } = a r^{ t – 1 } + a r^t + a r^{ t + 1 }\)
\( \vdots \)
\(S_n = a r^{ t – 1 } + a r^t + a r^{ t + 1 } + \cdots + a r^{ t + j – 1 }\)

A formula for \( S_n\) by using the following cleverness; Factor \( r^t \) from the \( n \)th partial sum yielding

$$S_n = r^t \left( a + a r^t + a r^{ t + 1 } + \cdots + a r^{ j – 1 } \right)$$

which is equal to

$$ S_n = r^t \left( \sum_{k=1}^{ j } ar^{k – 1} \right) = \frac{ r^t \left( a – a r^{ j } \right)}{ 1 – r } $$

Integration by Partial Fractions Problems

This page contains solutions to the following problems

$$\int \! \frac{1}{x \left( M – x \right) } \, dx$$
$$\int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx$$
$$\int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx$$

\( \displaystyle \int \! \frac{1}{x \left( M – x \right) } \ dx \)

Begin by decomposing the integrand.

$$ \frac{1}{x \left( M – x \right) } = \frac{A}{ x } + \frac{B}{ M – x } $$

Multiply through by \( x \left( M – x \right) \)

$$ 1 = A \left( M – x \right) + B x $$

If \( x = M \) then \( B = \frac{1}{ M } \)

If \( x = 0 \) then \( A = \frac{1}{ M } \)

therefore,

$$ \int \! \frac{1}{x \left( M – x \right) } \, dx = \frac{1}{M} \int \! \frac{1}{ x } \, dx + \frac{1}{M} \int \! \frac{1}{ M – x } \, dx $$
$$ \int \! \frac{1}{x \left( M – x \right) } = \frac{1}{M} \ln \left| x \right| – \frac{1}{M} \ln \left| M – x \right| + C = \frac{1}{M} \ln \left| \frac{x}{M – x} \right| + C$$

\( \displaystyle \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx \)

Decompose the integrand

$$ \frac{1}{ \left( ax + b \right) \left( cx + d \right) } = \frac{A}{ \left( ax + b \right)} \frac{B}{ \left( cx + d \right) }$$

Multiply through by \( \left( ax + b \right) \left( cx + d \right) \)

$$ 1 = A \left( cx + d \right) + B \left( ax + b \right) $$

If \( x = – \frac{d}{c} \), then \( 1 = B \left( a \left( -\frac{d}{c} \right) + b \right) \) yielding

$$ B = \frac{c}{ bc – ad } $$

Simimlarly,

If \( x = – \frac{b}{a} \), then \( 1 = A \left( c \left( -\frac{b}{a} \right) + d \right) \) and

$$ A = \frac{a}{ ad – bc } $$

substituting

$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx = \frac{a}{ ad – bc } \int \! \frac{1}{ \left( ax + b \right) } \, dx + \frac{c}{ bc – ad } \int \! \frac{1}{ \left( cx + d \right) } \, dx $$

\( p = a x + b \ \ \ \ \ \ dp = a \ dx \)

\( q = c x + d \ \ \ \ \ \ dq = c \ dx \)

$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx = \frac{a}{ ad – bc } \frac{1}{a} \int \! \frac{1}{ p } \, dx + \frac{c}{ bc – ad } \frac{1}{c} \int \! \frac{1}{ q } \, dx $$
$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{1}{ ad – bc } \ln \left( ax + b \right) + \frac{1}{bc – ad} \ln \left( c x + d \right) + C = \frac{1}{ad-bc} \ln \left( \frac{ a x + b }{ c x + d } \right) + C $$

\( \displaystyle \int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx \)

\( \frac{ k x + l }{ \left( ax + b \right) \left( cx + d \right) } = \frac{A}{ \left( ax + b \right)} \frac{B}{ \left( cx + d \right) }\)

\( k x + l = A \left( cx + d \right) + B \left( ax + b \right) \)

\( x = – \frac{d}{c} \)

\( k \left( – \frac{d}{c} \right) + l = B \left( a \left( -\frac{d}{c} \right) + b \right) \)

\( B = \frac{ c l – d k }{ bc – ad } \)

\( x = – \frac{b}{a} \)

\( k \left( – \frac{b}{a} \right) + l = A \left( c \left( -\frac{b}{a} \right) + d \right) \)

\( A = \frac{ a l – k b }{ ad – bc } \)

\( \int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{ a l – k b }{ ad – bc } \int \! \frac{1}{ a x + b } \ dx + \frac{ c l – d k }{ bc – ad } \int \! \frac{1}{ c x + d } \ dx \)

\( p = a x + b \ \ \ \ \ \ dp = a \ dx \)

\( q = c x + d \ \ \ \ \ \ dq = c \ dx \)

\( \int \! \frac{ k x + l }{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{a l – k b}{ ad – bc } \frac{1}{a} \int \! \frac{1}{ p } \ dx + \frac{ c l – d k }{ bc – ad } \frac{1}{c} \int \! \frac{1}{ q } \ dx \)

\( \int \! \frac{ k x + l}{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{a l – k b}{ a^2 d – abc } \ln \! \left( ax + b \right) + \frac{ c l – d k}{bc^2 – acd} \ln \! \left( c x + d \right) + C = \frac{ a l – k b}{ a^2 d – abc } \ln \! \left( ax + b \right) + \frac{ d k – c l }{acd – bc^2} \ln \! \left( c x + d \right) + C \)