19 Integration

At one level integration may be seen as the inverse process of differentiation. We would write:

$I = \int f'(x) dx = F(x) + C$

The $+C$ is called the constant of integration. If you differentiate a constant you get zero. If you integrate an expression there may have been a constant value so we add $+C$ until we can evaluate the constant by using the boundary conditions.

19.1 Table of Common Standard Integrals

In the following table the left hand column contains functions of $x$ ($y=f(x)$). The right hand column contains the integrals of the functions with respect to $x$ ($ \int f(x) dx=F(x)$).


$y=f(x)$    $ \int f(x) = F(x)$
$x^n$$ \frac{x^{n+1}}{n+1}+C$
$x^{-1}$$ ln(x)+C$ for $x>0$
 $ ln(-x)+C$ for $x<0$
$cos(ax)$$sin(ax)/a+C$
$sin(ax)$$-cos(ax)/a+C$
$tan(ax)$$ln(sec(ax))/a+C$
$e^{ax}$$e^{ax}/a+C$
$ln(ax)$$x(ln(ax)-1)+C$

Example 19.1

$I=\int 2x^3 dx$

$I=\frac{2x^4}{4}+C$

$I=\frac{x^4}{2}+C$

Sanity Check
$\frac{d}{dx}(\frac{x^4}{2}+C)$=$4 \times \frac{x^3}{2}$
 =$2x^3$

Example 19.2

$I=\int cos(3x) dx$

$I=sin(3x)/3+C$

Sanity Check
$\frac{d}{dx}(sin(3x)/3+C)$=$3 \times cos(3x)/3$
 =$cos(3x)$

Example 19.3

$I=\int sin(4x) dx$

$I=-cos(4x)/4+C$

Sanity Check
$\frac{d}{dx}(-cos(4x)/4+C)$=$4 \times sin(4x)/4$
 =$sin(4x)$

19.2 Integrating Polynomial Expressions

Polynomial expressions may be integrated term by term. If we have an expression like $3x^2-2x+5$ the integral would be:

$I$$=$\( \int (3x^2-2x+5)dx \)
 $=$$$\frac{3x^3}{3}-\frac{2x^2}{2}+5x+C$$
 $=$\( x^3-x^2+5x+C \)

Example 19.4

$I=\int (4x-3)^3 dx$

$I=\int (64x^3-144x^2+108x-27) dx$

$I=64x^4/4-144x^3/3+108x^2/2-27x+C$

$I=16x^4-48x^3+54x^2-27x+C$

Sanity Check
$\frac{d}{dx}(16x^4-48x^3+54x^2-27x+C)$=$4 \times 16x^3 - 3 \times 48x^2 + 2 \times 54x - 27$
 =$64x^3-144x^2+108x-27$
 =$(4x-3)^3$

19.3 Integration by Substitution

Expressions that include a function of a linear factor may be integrated using a substitution.

If we want to integrate $(3x+4)^5$ we let $3x+4=u$.

Differentiating we get $3dx=du$ so $dx=du/3$.

Substituting into the integral we get

$I$$=$\( \int (3x+4)^5dx \)
 $=$$$\frac{1}{3}\int u^5du$$
 $=$$$ \frac{1}{3} \frac{u^6}{6}+C $$
 $=$$$ \frac{(3x+4)^6}{18}+C $$

Example 19.5

$I=\int (4x-3)^3 dx$

Let $u = 4x-3$ so $du = 4dx$ or $dx = \frac{1}{4}du$

$I=\frac{1}{4}\int u^3 du$

$I=\frac{1}{4}(u^4/4)+C$

Substituting $u = 4x-3$

$I=\frac{1}{16}(4x-3)^4+C$

Sanity Check
$\frac{d}{dx}(\frac{1}{16}(4x-3)^4+C)$=$\frac{4}{16}(4x-3)^3 \times 4$
 =$(4x-3)^3$

Example 19.6

$I=\int \frac{4}{2x-3} dx$

Let $u = 2x-3$ so $du = 2dx$ or $dx = \frac{1}{2}du$

$I=4\int \frac{1}{2u} du$

$I=2\int \frac{1}{u} du$

$I=2 ln(u)+C$

Substituting $u = 2x-3$

$I=2 ln(2x-3)+C$

Sanity Check
$\frac{d}{dx}(2ln(2x-3))$=$2 \times \frac{1}{2x-3} \times 2$
 =$\frac{4}{2x-3}$

19.4 Integration by Partial Fractions

An expression like $$I=\int \frac{2x+23}{2x^2+x-10}dx$$ cannot be integrated using the table of standard integrals but we may be able to simplify the expression using partial fractions. First we need to see whether we can factorise the denominator.

You can probably see that $2x^2+x-10=(2x+5)(x-2)$ so we could write

$$\frac{2x+23}{2x^2+x-10}$$$=$$$\frac{A}{2x+5}+\frac{B}{x-2}$$
where $A$ and $B$ are unknown constants.
Putting the right hand side over a common denominator we get
 $=$$$\frac{A(x-2)+B(2x+5)}{(2x+5)(x-2)}$$
so $2x+23$$=$$A(x-2)+B(2x+5)$
 
If we let $x=2$ we get $4+23=B(4+5)$ so $B=3$
 
Now let $x=-5/2$ then $-5+23=A(-5/2-2)$ so $A=4$
We have found values for $A$ and $B$ so we can write
$$\int \frac{2x+23}{2x^2+x-10}dx$$$=$$$\int \frac{4}{2x+5}dx+\int \frac{3}{x-2}dx$$
Integrating we get $I$$=$$4 ln(2x+5)/2+3 ln(x-2)+C$
 
 $=$$2 ln(2x+5)+3 ln(x-2)+C$

19.4 Integration by Parts

We know

$$\frac{d}{dx}(uv)=v\frac{du}{dx}+u\frac{dv}{dx}$$

If we integrate this equation we get

$$uv=\int v\frac{du}{dx}dx+\int u\frac{dv}{dx}dx$$

Rearranging this we get

$$\int v\frac{du}{dx}dx=uv-\int u\frac{dv}{dx}dx$$

If we have to integrate an expression like $x sin(x)$ we could let $v=x$ and $du=sin(x)$

which means $dv=dx$ and $u=-cos(x)$ so we can write

$\int x sin(x) dx$$=$$-x cos(x)-\int (-cos(x))dx$
 $=$$-x cos(x)+sin(x) +C$
 $=$$sin(x)-x cos(x)+C$