Category Archives: Solved Problems

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\( x = – \frac{d}{c} \)

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

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

\( x = – \frac{b}{a} \)

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

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

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

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

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

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

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

Integration by Parts Problems

$$ \int \! u \ dv = uv \ – \int \! v \ du$$

\( \int \! \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = dx \ \ \ \ \ \ v = x \)

\( \int \! \ln x \ dx = x \ln x – \int \! x \frac{1}{x} \ dx \)

\( \int \! \ln x \ dx = x \ln x – x + C \)

 

\( \int \! x \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = x \ dx \ \ \ \ \ \ v = \frac{x^2}{2}\)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \int \! \frac{x^2}{2} \frac{1}{x} \ dx \)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C \)

 

\( \int \! \sin^{-1} x \ dx \)

\( u = \sin^{-1} x \ \ \ \ \ \ du = \frac{1}{\sqrt{1 – x^2}} \ dx\)

\( dv = dx \ \ \ \ \ \ v = x \)

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

\( \int \! \frac{x}{ \sqrt{1 – x^2 }} \ dx\)

\( t = 1 – x^2 \ \ \ \ \ \ – \frac{1}{2} \ dt = x \ dx\)

\( – \int \! \frac{1}{ 2 \sqrt{ t }} \ dt = – \sqrt{ t } + C\)

\( \int \! \sin^{-1} x \ dx = x \sin^{-1} x – \int \! x \frac{1}{ \sqrt{1 – x^2 }} \ dx = x \sin^{-1} x + \sqrt{ 1 – x^2 } + C \)

 

\( \int \! x e^{ x } \ dx \)

\( u = x \ \ \ \ \ \ du = dx \)

\( dv = e^{x} \ dx \ \ \ \ \ \ v = e^{x}\)

\( \int \! x e^{ x } \ dx = x e^{x} – \int \! e^{x} \ dx = x e^{x} – e^{x} + C\)

 

\( \int \! e^{ \sqrt{x} } \ dx \)

\( t = \sqrt{x} \ \ \ \ \ \ dt = \frac{1}{2 \sqrt{ x }} \ dx\)

\( 2 t \ dt = dx \)

\( \int \! 2 t e^{ t } \ dt \)

\( u = 2 t \ \ \ \ \ \ du = 2 \ dt \)

\( dv = e^{t} \ dt \ \ \ \ \ \ v = e^{t}\)

\( \int \! 2 t e^{ t } \ dt = 2 t e^{t} – \int \! 2 e^{t} \ dt = 2 t e^{t} – 2 e^{t} + K\)

\( \int \! e^{ \sqrt{x} } \ dx = 2 \sqrt{ x } e^{ \sqrt{ x } } -2 e^{ \sqrt{ x } } + C\)

 

\( \int \! e^{ x } \sin x \ dx \)

\( u = \sin x \ \ \ \ \ \ dv = e^{ x } \ dx\)

\( du = \cos x \ dx \ \ \ \ \ \ v = e^{ x } \)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \int \! e^{ x } \cos x \ dx \)

\( s = \cos x \ \ \ \ \ \ ds = – \sin x \ dx \)

\( dt = e^{x} \ dx \ \ \ \ \ \ t = e^{x}\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \left( e^{ x } \cos x – \int \! – e^{ x } \sin x \ dx \right)\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – e^{ x } \cos x – \int \! e^{ x } \sin x \ dx \)

\( \int \! e^{ x } \sin x \ dx = \frac{1}{2} \left( e^{ x } \sin x – e^{ x } \cos x \right) + C \)

Differential Equations: Worked Examples Using Integrating Factor

$$\frac{dy}{dx} – 2y = 3x + 1$$
$$e^{ \int \! -2 \ dx } = e^{-2x}$$
$$e^{-2x} \frac{dy}{dx} – 2 e^{-2x} y = e^{-2x} \left( 3x + 1 \right)$$
$$\frac{d}{dx} \left( e^{-2x} y \right) = 3 x e^{-2x} + e^{-2x}$$
$$e^{-2x} y + C = \int \! 3 x e^{-2x} \ dx + \int \! e^{-2x} \ dx$$
$$e^{-2x} y = \int \! 3 x e^{-2x} \ dx – \frac{1}{2}e^{-2x} + K \ \ where \ \ K = \left( C_0 – C \right)$$

\( \int \! 3 x e^{-2x} \ dx \)

\( u = 3x \)

\( du = 3 \ dx \)

\( dv = e^{-2x} \ dx \)

\( v = – \frac{1}{2}e^{-2x}\)

\( \int \! 3 x e^{-2x} \ dx = – \frac{3}{2}xe^{-2x} + \int \! \frac{3}{2}e^{-2x} \ dx \)

\( \int \! 3 x e^{-2x} \ dx = – \frac{3}{2}xe^{-2x} – \frac{3}{4}e^{-2x} + C_1 \)

$$e^{-2x} y = – \frac{3}{2}xe^{-2x} – \frac{5}{4}e^{-2x} + C_2 \ \ where \ \ C_2 = \left( K + C_1 \right)$$
$$y \left( x \right) = – \frac{3}{2}x – \frac{5}{4} + C_2e^{2x}$$

 

$$\frac{dy}{dx} + \cos \left(x \right) y = \cos x$$
$$\mu = e^{ \int \! \cos x \ dx } = e^{ \sin x }$$
$$e^{ \sin x } \frac{dy}{dx} + e^{ \sin x } \cos \left(x \right) y = e^{ \sin x } \cos x$$
$$\frac{d}{dx} \left( e^{ \sin x } y \right) = e^{ \sin x } \cos x$$
$$\left( e^{ \sin x } y \right) = \int \! e^{ \sin x } \cos x \ dx + C$$

\( v = \sin x \)

\( dv = \cos x \ dx \)

\( \int \! e^v \ dv = e^v + K = e^{ \sin x } + C_1\)

$$\left( e^{ \sin x } y \right) = e^{ \sin x } + C_1 + C = e^{ \sin x } + C_2$$
$$y \left( x \right) = 1 + C_2 e^{ – \sin x }$$

Using the Integrating Factor

Let \( g \) and \( h \) be continuous functions. The form of the following first order linear differential equation is solvable using an integrating factor.

$$\frac{dy}{dx} + g y = h$$

The left side of the equation almost looks like the result of the product rule. For example, if \(j\left(x \right) = \mu \left(x \right) y\left(x \right)\) then

$$\frac{dj}{dx} = \mu \frac{dy}{dx} + \frac{d \mu }{dx} y$$

Comparing

\( \frac{dy}{dx} + g y \)         to         \( \mu \frac{dy}{dx} + \frac{d \mu }{dx} y \)

suggests that if there exists a function \( \mu \left(x \right) \) such that

$$\frac{d \mu}{dx} = \mu g$$

then multiplying both sides of the original differential equation by \( \mu \left(x \right) \) yields

$$\mu \frac{dy}{dx} + \mu g y = \mu h$$

and since \(\frac{d \mu}{dx} = \mu g \)

$$\mu \frac{dy}{dx} + \mu g y = \frac{d \mu}{dx} y + \mu \frac{dy}{dx} = \frac{d}{dx} \left( \mu y \right)$$

which yields the integrable equation

$$\frac{d}{dx} \left( \mu y \right) = \mu h$$

Integrating yields a formula the for the solution to the differential equation in question

$$y = \frac{1}{\mu} \left( \int \! \mu h \ dx + C \right)$$

If this formula is to be of any practical use \( \mu \) must be determined. Recall that

$$\frac{d \mu}{dx} = \mu g$$

which is a separable differential equation. Therefore,

$$\mu = e^{ \int \! g\left(x \right) \ dx + K} = A e^{ \int \! g\left(x \right) \ dx}$$

Since the arbitrary constant will not affect the final answer it is convention to let \( A = 1 \). \( \mu \) is called the integrating factor. Substituting \( \mu \) into the general solution yields

$$y\left(x \right) = e^{ -\int \! g\left(x \right) \ dx} \left( \int \! e^{ \int \! g\left(x \right) \ dx} h\left(x \right) \ dx + C \right)$$