Rules of Integration

dy/dx = a.xⁿ

dy = ∫a.xⁿ.dx
y = a.xⁿ⁺¹/n+1

dy/dx = u + v

dy = ∫u.dx + ∫v.dx

Trigonometric Functions

A fairly straight-forward example of this integration procedure is provided below; Integrate ∫Sin³(x)/Cos²(x).dx

∫Sin(x).Sin²(x)/Cos²(x).dx
{as Sin²(x) = 1-Cos²(x) see Trig Functions}
∫Sin(x).(1-Cos²(x))/Cos²(x).dx
∫Sin(x)/Cos²(x).dx - ∫Sin(x).Cos²(x)/Cos²(x).dx
∫Sin(x)/Cos²(x).dx - ∫Sin(x).dx
{as Sin(x)/Cos(x) = Tan(x) see Trig Functions}
∫Tan(x)/Cos(x).dx - ∫Sin(x).dx
{as 1/Cos(x) = Sec(x) see Trig Functions}
∫Tan(x).Sec(x).dx - ∫Sin(x).dx
{as ∫Tan(x).Sec(x).dx = Sec(x) see Standard Integrals}
Sec(x) + Cos(x)

Algebraic Substitution

A fairly straight-forward example of this integration procedure is provided below; Integrate ∫x/(x+3)^½.dx

If you set; u = (x+3)^½; then u² = x+3 or x = u²-3
{and if x = u²-3 then dx = 2u.du}

Substituting:
∫x/(x+3)^½.dx = ∫(u²-3) / u . 2u.du
... = 2∫(u²-3).du
... = 2∫u².du - 2∫3.du

... = 2.u³/3 - 2.3.u
... = 2.(u³/3 - 3u)
... = 2u.(u²/3 - 3)
... = ⅔u.(u² - 9)

Reverse Substitution:
{because u = (x+3)^½ and u² = x+3}
... = ⅔(x+3)^½.(x+3 - 9)
... = ⅔(x+3)^½.(x-6)

Trigonometric Substitutions
(typical)

With practice you will get to see the best way to reorganise your formulas for ease of integration. To give you some idea of how to do this, see the following worked examples:

Integrate ∫Sin²(x).dx
note: Sin²(x) = Sin(x).Sin(x) = ½(Cos(x-x)-Cos(x+x))
... = ½(1-Cos(2x)) {Cos(x-x) = Cos(0) = 1}
Therefore Sin²(x) = ½(1-Cos(2x))
∫Sin²(x).dx = ∫½(1-Cos(2x)).dx = ½∫dx - ½∫Cos(2x).dx
... = ½.x - ½.Sin(2x)/2
x/2 - Sin(2x)/4

Integrate ∫Sin(x).Cos(x).dx
note: Sin(x).Cos(x) = ½(Sin(x+x)+Sin(x-x))
... = ½(Sin(2x)+0) {Sin(x-x) = Sin(0) = 0}
Therefore Sin(x).Cos(x) = ½.Sin(2x)
∫Sin(x).Cos(x).dx = ∫½.Sin(2x).dx = ½∫Sin(2x).dx
... = -½.Cos(2x)/2
-Cos(2x)/4

By Parts

This method is based upon a reverse procedure for the differentiation 'Product Rule':
i.e. dy/dx = u.dv/dx + v.du/dx

If y = uv
d(uv)/dx = u.dv/dx + v.du/dx
uv = ∫u.dv.dx/dx + ∫v.du.dx/dx
uv = ∫u.dv + ∫v.du
∫u.dv = uv - ∫v.du #

Example: Integrate ∫3x.Sin(x).dx
Set u = 3x and dv = Sin(x).dx
then du = 3.dx and v = -Cos(x)

As ∫u.dv = uv - ∫v.du {see above #}
∫u.dv = 3x.-Cos(x) - ∫-Cos(x).3.dx
∫u.dv = 3∫Cos(x).dx - 3x.Cos(x)

∫3x.Sin(x).dx = 3∫Cos(x).dx - 3x.Cos(x)
Integrating 3∫Cos(x).dx = 3.Sin(x) ...

∫3x.Sin(x).dx = 3.Sin(x) - 3x.Cos(x)

e.g.: ℓ = ʃ(1 + b²/(a²-a⁴/x²))⁰˙⁵ is a painful integral to solve by normal integration

Simpson's Rule may be applied as follows:

1) replace (1 + b²/(a²-a⁴/x²))⁰˙⁵ with ƒ(x)

2) select points x₁ and x₂ on your curve between which you need an arc length (ℓ)

3) divide this distance (x₂-x₁) by any even number (n) thus: δx = (x₂-x₁)/n
the larger the number (n) the more accurate the result (see Examples 1 & 2 below)

4) generate the following formula
ℓ = δx/3 . Σ[ƒ(x₁) + 4.ƒ(x₁+δx) + 2.ƒ(x₁+2.δx) + 4.ƒ(x₁+3.δx) + 2.ƒ(x₁+4.δx).......
+ 4.ƒ(x₁+(n-1).δx) + ƒ(x₁+n.δx)]
note: x₁+n.δx = x₂

5) replace each function term with the formula thus: e.g.
... + 2.ƒ(x₁+4.δx) + ... = ... + 2.(1 + b²/(a²-a⁴/(x₁+4.δx)²))⁰˙⁵ + ...

So Simpson's version of the formula now looks like this:
ℓ = δx/3 . Σ[(1 + b²/(a²-a⁴/(x₁)²))⁰˙⁵ + 4.(1 + b²/(a²-a⁴/(x₁+δx)²))⁰˙⁵ +
2.(1 + b²/(a²-a⁴/(x₁+2.δx)²))⁰˙⁵ + 4.(1 + b²/(a²-a⁴/(x₁+3.δx)²))⁰˙⁵ +
2.(1 + b²/(a²-a⁴/(x₁+4.δx)²))⁰˙⁵...... + 4.(1 + b²/(a²-a⁴/(x₁+(n-1).δx)²))⁰˙⁵ +
(1 + b²/(a²-a⁴/(x₁+n.δx)²))⁰˙⁵]

which is easily solved

The following two examples give you some idea of the errors you can expect with Simpson's Rule, dependent upon the number of terms (n) you use for your calculation

Example 1: ∫1/(x⁰˙⁵ + 2).dx = 2.(x⁰˙⁵ - 2.Ln(x⁰˙⁵ + 2))
where f(x) = 1/(x⁰˙⁵ + 2)
if x₁ = 4.7 and x₂ = 7.9; 2.(x⁰˙⁵ - 2.Ln(x⁰˙⁵ + 2)) = 0.71182137680140

Example 2: ∫(b.x + c.x²)⁰˙⁵ / x³.dx = -2.((b.x + c.x²)³)⁰˙⁵ / 3.b.x³
where f(x) = (b.x + c.x²)⁰˙⁵ / x³
if x₁ = 2 and x₂ = 4.73; -2.((b.x + c.x²)³)⁰˙⁵ / 3.b.x³ = 0.6172678975456420