• The solution to Newton's GEXACT VALUE & FORMULA
  • The theory controlling planetary spinTHE MATHEMATICAL LAW
  • The pressure at the centre of a massEARTH'S CORE PRESSURE (calculation procedure)
  • Proof of the non-exitence of Dark MatterDOES NOT EXIST
  • The atom as Newton and Coulomb describe itNO NEED FOR A UNIFICATION THEORY
The solution to Newton's G1 The theory controlling planetary spin2 Pressure at the centre of the Earth3 Proof of the non-exitence of Dark Matter4 The atom as Newton describes it5
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Integration of Algebraic and Trigonometric Functions

The following table contains integrated examples of basic algebraic and trigonometric formulas.
Ln means natural logarithm

∫dx

x

∫xn.dx

xn+1 / (n+1)

∫axn.dx

a . xn+1 / (n+1)

∫(axn + b).dx
= ∫axn.dx + ∫b.dx

a.xn+1 / (n+1) + b.x

∫(ax + b)n.dx

(ax+b)n+1 / a(n+1)

∫dx / (ax + b)
= 1/a . ∫a.dx / (ax + b)

1/a . Ln(ax+b)

∫1/x . dx

Ln(x)

∫1/(x + b)½ . dx

2(x+b)½

∫1/(ax + b)½ . dx

2(ax+b)½ / a

∫1/(x² - a²) . dx

-Acoth(x/a) / a
or
Ln[(x-a)/(x+a)] / 2a

∫1/(a² - x²) . dx

Atanh(x/a) / a
or
Ln[(a+x)/(a-x)] / 2a

∫1/(a² + x²) . dx

Atan(x/a) / a

∫(x² + a²)½ . dx

½x(x² + a²)½ + ½a² . Asinh(x/a)
or
½x(x² + a²)½ + ½a² . Ln([x+(x² + a²)½] / a)

∫ƒ'(x)/ƒ(x) . dx = Ln(ƒ(x))

Note: If the numerator = the differential of the denominator then the inverse of the denominator is the logₑ of the denominator.
So multiply the equation by the differential of the denominator and 'Logₑ' the result

d(u.v) / dx

u.v = ∫u.dv/dx.dx +∫v.du/dx.dx = ∫u.dv +∫v.du
∫u.dv = u.v - ∫v.du

∫ax . dx

ax . loga(e)

∫ex . dx

ex

∫Sin(x) . dx

–Cos(x)

∫Cos(x) . dx

Sin(x)

∫Tan(x) . dx

–Ln(Cos(x)), or
Ln(Sec(x))

∫Cot(x) . dx

Ln(Sin(x))

∫Sec(x) . dx

Ln(Tan(¼π + ½x))

∫Cosec(x) . dx

Ln(Tan(½x))

∫Sinh(x) . dx

Cosh(x)

∫Cosh(x) . dx

Sinh(x)

∫Tanh(x) . dx

Ln(Cosh(x))

∫Coth(x) . dx

Ln(Sinh(x))

∫Sin(ax) . dx

–Cos(ax) / a

∫Sin(ax + b) . dx

–Cos(ax + b) / a

∫Cos(ax) . dx

Sin(ax) / a

∫Cos(ax + b) . dx

Sin(ax + b) / a

∫Tan(ax) . dx

Ln(Sec(ax)) / a

∫Sinh(ax) . dx

Cosh(ax) / a

∫Cosh(ax) . dx

Sinh(ax) / a

∫Sin(x).Cos(x) . dx

-¼Cos(2x)

∫Sec(x).Tan(x) . dx

Sec(x)

∫Csc(x).Cot(x) . dx

–Csc(x)

∫1 / (a² – x²)½ . dx

Asin(x/a), or
–Acos(x/a)

∫1 / (a² + x²) . dx

Asec(x/a) / a, or
–Acsc(x/a) / a

∫1 / x(x² – a²)½ . dx

Asec(x/a) / a, or
–Acsc(x/a) / a

∫1 / (x² + a²)½ . dx

Asinh(x/a), or
Ln(x+(x²+a²)½ / a)

∫1 / (x² – a²)½ . dx

Acosh(x/a), or
Ln(x+(x²–a²)½ / a)

∫1 / (a² – x²) . dx

Atanh(x/a) / a, or
Ln((a+x)/(a–x)) / 2a

∫1 / (x² – a²) . dx

–Acoth(x/a) / a, or
Ln((a–x)/(a+x)) / 2a

∫1 / x(a² – x²)½ . dx

–Asech(x/a) / a, or
–Ln((a + (a²–x²)½) / x) / a

∫1 / x(a² + x²)½ . dx

–Acsch(x/a) / a, or
–Ln((a + (a²+x²)½) / x) / a

∫Sin²(x) . dx

½(x – ½.Sin(2x))

∫Cos²(x) . dx

½(x + ½.Sin(2x))

∫Tan²(x) . dx

Tan(x) – x

∫Csc²(x) .dx

–Cot(x)

∫Sec²(x) . dx

Tan(x)

∫Cot²(x) . dx

–(Cot(x) + x)

∫(x² – a²)½ .dx

½.x(x²–a²)½ – a².Acosh(x/a)/2, or
½.x(x²–a²)½ – a²(logₑ((x+(x²–a²)½ / a) / 2

∫(x² + a²)½ .dx

½.x(x²+a²)½ + a².Asinh(x/a)/2, or
½x(x²+a²)½ + a²(logₑ((x+(x²+a²)½ / a) / 2

∫(a² – x²)½ .dx

½.a².Asin(x/a) + ½.x(a² – x²)½

∫Sin²(ax)

½x – ¼Sin(2ax)/a

∫x.Sin(ax).dx

Sin(ax)/a² – x.Cos(ax)/a

∫x².Sin(ax)

-x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2Cos(ax)/a³

∫x².Sin²(ax)

x³/6 – ¼.x².Sin(2ax)/a – ¼x.Cos(2ax)/a² + ⅛Sin(2ax)/a³

∫x³.Sin(ax)

-x³.Cos(ax)/a + 3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴

∫Cos²(ax)

¼Sin(2ax)/a + ½x

∫x.Cos(ax).dx

x.Sin(ax)/a + Cos(ax)/a²

∫x².Cos(ax)

x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³

∫x².Cos²(ax)

¼.x².Sin(2ax)/a + x³/6 + x.Cos(2ax) / 4a² – ⅛Sin(2ax)/a³

∫x³.Cos(ax)

x³.Sin(ax)/a + 3x².Cos(ax)/a² – 6.x.Sin(ax)/a³ – 6.Cos(ax)/a⁴

∫Sin(x).Cos(x)

-¼.Cos(2x)

Worked Examples

The following table contains a number of examples worked through by CalQlata engineers from time to time.
The table may not yet be complete but will be eventually. We are adding new integral workings as we resolve them.

Note: there are a number of different ways to integrate these formulas, we have simply listed the methods we have used.

Typical Integration by Substitution:
Problem: ∫(a + b.x²)⁰˙⁵ . dx

set: m = √a; n = √b; x = m/n . Tan(θ)     {i.e. θ = Atan[x.n/m]}
note: Sec²(θ) = 1+Tan²(θ)

∫(m² + n².x²)⁰˙⁵ . dx
     = ∫(m² + .m²/ . Tan²[θ])⁰˙⁵ . dθ
     = ∫(m² + m² . Tan²[θ])⁰˙⁵ . dθ
     = ∫(m².(1 + Tan²[θ]))⁰˙⁵ . dθ
     = ∫(m².Sec²[θ])⁰˙⁵ . dθ
     = ∫m.Sec[θ] . dθ
     = m∫Sec[θ] . dθ
     = m . Ln(Tan[¼π + ½θ])     {see ∫Sec[x].dx above}

substitute back:
for x: m . Ln(Tan[¼π + ½{Atan[x.n/m]}])
for a & b: √a . Ln(Tan[¼π + ½{Atan[x.√b/√a]}])

∫(a + b.x²)⁰˙⁵ . dx = √a . Ln(Tan[¼π + ½.Atan[x.√(b/a)]])

∫Sin²(x).dx

Sin²(x) = Sin(x).Sin(x)
     = ½(Cos(x–x) – Cos(x+x))
     = ½(Cos(0) – Cos(2x))
     = ½(1 – Cos(2x))
     = ½ – ½Cos(2x)

∫Sin²(x) = ∫(½ – ½Cos(2x)).dx
     = ∫½.dx – ∫½Cos(2x).dx
     = ½∫dx – ½∫Cos(2x).dx
     = ½.x – ½.Sin(2x)/2
∫Sin²(x) = ½x – ¼Sin(2x)

∫Sin²(ax).dx

Sin²(ax) = Sin(ax).Sin(ax)
     = ½(Cos(ax–ax) – Cos(ax+ax))
     = ½(Cos(0) – Cos(2ax))
     = ½(1 – Cos(2ax))
     = ½ – ½Cos(2ax)

∫Sin²(ax) = ∫(½ – ½Cos(ax)).dx
     = ∫½.dx – ∫½Cos(2ax).dx
     = ½∫dx – ½∫Cos(2ax).dx
     = ½x – ½Sin(2ax)/2a
∫Sin²(ax) = ½x – ¼Sin(2ax)/a

∫x.Sin(ax).dx
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x; dv = Sin(ax); du = dx; v = -Cos(ax)/a

∫x.Sin(ax).dx = x.-Cos(ax)/a – ∫-Cos(ax)/a.dx
     = -x.Cos(ax)/a + 1/a∫Cos(ax).dx
     = -x.Cos(ax)/a + 1/a.Sin(ax)/a
     = -x.Cos(ax)/a + Sin(ax)/a²
∫x.Sin(ax).dx = Sin(ax)/a² – x.Cos(ax)/a

∫x².Sin(ax)
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x²; dv = Sin(ax); du = 2x.dx; v = -Cos(ax)/a

∫x².Sin(ax) = x².-Cos(ax)/a - ∫-Cos(ax)/a . 2x.dx
∫x².Sin(ax) = -x².Cos(ax)/a + 2/a∫x.Cos(ax).dx

∫x.Cos(ax).dx
u = x; dv = Cos(ax); du =dx; v = Sin(ax)/a
∫x.Cos(ax).dx = x . Sin(ax)/a – ∫Sin(ax)/a . dx
     = x . Sin(ax)/a – 1/a∫Sin(ax).dx
     = x . Sin(ax)/a – 1/a-Cos(ax)/a.dx
     = x.Sin(ax)/a + Cos(ax)/a/a
∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a²

∫x².Sin(ax) = -x².Cos(ax)/a + 2/a . (x.Sin(ax)/a + Cos(ax)/a²)
     = -x².Cos(ax)/a + (2/a . x.Sin(ax)/a + 2/a . Cos(ax)/a²)
     = -x².Cos(ax)/a + (2x.Sin(ax)/a² + 2Cos(ax)/a³)
∫x².Sin(ax) = 2Cos(ax)/a³ + 2x.Sin(ax)/a² – x².Cos(ax)/a

∫x².Sin²(ax)

Sin²(ax) = Sin(ax).Sin(ax)
     = ½(Cos(ax–ax) – Cos(ax+ax))
     = ½(Cos(0) – Cos(2ax))
     = ½(1 – Cos(2ax))
Sin²(ax) = ½ – ½Cos(2ax)

(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x²; dv = ½ – ½Cos(2ax); du = 2x.dx; v = ½x – ¼Sin(2ax)/a
∫x².Sin²(ax) = x².(½x – ¼.Sin(2ax)/a) – ∫(½x – ¼.Sin(2ax)/a) . 2x.dx
     = ½x³ – ¼.x².Sin(2ax)/a – ∫(x² – ½.x.Sin(2ax)/a).dx
     = ½x³ – ¼.x².Sin(2ax)/a – ∫x².dx + ∫½.x.Sin(2ax)/a.dx
     = ½x³ – ¼.x².Sin(2ax)/a – ∫x².dx + 1 / 2a∫x.Sin(2ax).dx
     = ½x³ – ¼.x².Sin(2ax)/a – ⅓x³ + 1 / 2a∫x.Sin(2ax).dx
∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a + 1 / 2a∫x.Sin(2ax).dx

∫x.Sin(2ax).dx
u = x; dv = Sin(2ax); du = dx; v = -Cos(2ax)/2a
∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a – ∫-Cos(2ax) / 2a . dx
     = -x.Cos(2ax) / 2a + 1 / 2a∫Cos(2ax) . dx
     = -x.Cos(2ax) / 2a + 1 / 2a.Sin(2ax) / 2a
∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a + Sin(2ax) / 4a²

∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a + 1 / 2a . (-x.Cos(2ax) / 2a + Sin(2ax) / 4a²)
     = x³/6 – ¼.x².Sin(2ax)/a + (-x.Cos(2ax) / 4a² + ⅛Sin(2ax)/a³)
∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a – ¼x.Cos(2ax)/a² + ⅛Sin(2ax)/a³

∫x³.Sin(ax)
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x³; dv = Sin(ax); du = 3.x².dx; v = -Cos(ax)/a
∫x³.Sin(ax) = x³.-Cos(ax)/a – ∫-Cos(ax)/a . 3x².dx
∫x³.Sin(ax) = -x³.Cos(ax)/a + 3/a∫x².Cos(ax).dx

∫x².Cos(ax).dx
u = x²; dv = Cos(ax); du =2x.dx; v = Sin(ax)/a
∫x².Cos(ax).dx = x².Sin(ax)/a – ∫Sin(ax)/a . 2x.dx
     = x².Sin(ax)/a – 2/a∫Sin(ax) . x.dx
∫x².Cos(ax).dx = x².Sin(ax)/a – 2/a∫x.Sin(ax).dx

∫x.Sin(ax).dx
u = x; dv = Sin(ax); du =dx; v = -Cos(ax)/a
∫x.Sin(ax).dx = x . -Cos(ax)/a – ∫-Cos(ax)/a . dx
     = -x.Cos(ax)/a + 1/a∫Cos(ax).dx
     = -x.Cos(ax)/a + Sin(ax)/a/a
∫x.Sin(ax).dx = -x.Cos(ax)/a + Sin(ax)/a²

∫x².Cos(ax).dx = x².Sin(ax)/a – 2/a . (-x.Cos(ax)/a + Sin(ax)/a²)
     = x².Sin(ax)/a – (2/a.-x.Cos(ax)/a + 2/aSin(ax)/a²)
     = x².Sin(ax)/a – (2.-x.Cos(ax)/a² + 2.Sin(ax)/a³)
∫x².Cos(ax).dx = x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³

∫x³.Sin(ax) = -x³.Cos(ax)/a + 3/a . (x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³)
     = -x³.Cos(ax)/a + (3/a . x².Sin(ax)/a + 3/a . 2.x.Cos(ax)/a² – 3/a . 2.Sin(ax)/a³)
     = -x³.Cos(ax)/a + (3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴)
∫x³.Sin(ax) = -x³.Cos(ax)/a + 3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴

∫Cos²(x).dx

Cos²(x) = Cos(x).Cos(x)
     = ½(Cos(x+x) + Cos(x-x))
     = ½(Cos(2x) + Cos(0))
     = ½(Cos(2x) + 1)
     = ½Cos(2x) + ½

∫Cos²(x) = ∫(½Cos(2x) + ½).dx
     = ∫½Cos(2x).dx + ∫½.dx
     = ½∫Cos(2x).dx + ½∫dx
     = ½.Sin(2x)/2 + ½.x
∫Cos²(x) = ¼Sin(2x) + ½x

∫Cos²(ax).dx

Cos²(ax) = Cos(ax).Cos(ax)
     = ½(Cos(ax+ax) + Cos(ax-ax))
     = ½(Cos(2ax) + Cos(0))
     = ½(Cos(2ax) + 1)
     = ½Cos(2ax) + ½

∫Cos²(ax) = ∫(½Cos(ax) + ½).dx
     = ∫½Cos(2ax).dx + ∫½.dx
     = ½∫Cos(2ax).dx + ½∫dx
     = ½Sin(2ax)/2a + ½x
∫Cos²(ax) = ¼Sin(2ax)/a + ½x

∫x.Cos(ax).dx
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x; dv = Cos(ax); du = dx; v = Sin(ax)/a

∫x.Cos(ax).dx = x.Sin(ax)/a – ∫Sin(ax)/a.dx
     = x.Sin(ax)/a – 1/a∫Sin(ax).dx
     = x.Sin(ax)/a – 1/a.-Cos(ax)/a
     = x.Sin(ax)/a + Cos(ax)/a²
∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a²

∫x².Cos(ax)
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x²; dv = Cos(ax); du =2x.dx; v = Sin(ax)/a
∫x².Cos(ax) = x².Sin(ax)/a – ∫Sin(ax)/a . 2x.dx
∫x².Cos(ax) = x².Sin(ax)/a – 2/a∫x.Sin(ax).dx

∫x.Sin(ax).dx
u = x; dv = Sin(ax); du =dx; v = -Cos(ax)/a
     = x . -Cos(ax)/a – ∫-Cos(ax)/a . dx
     = -x.Cos(ax)/a + 1/a∫Cos(ax).dx
     = -x.Cos(ax)/a + 1/a.Sin(ax)/a
∫x.Sin(ax).dx = -x.Cos(ax)/a + Sin(ax)/a²

∫x².Cos(ax) = x².Sin(ax)/a – 2/a . (-x.Cos(ax)/a + Sin(ax)/a²)
     = x².Sin(ax)/a – (2/a.-x.Cos(ax)/a + 2/a.Sin(ax)/a²)
     = x².Sin(ax)/a – (2.-x.Cos(ax)/a² + 2.Sin(ax)/a³)
∫x².Cos(ax) = x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³

∫x².Cos²(ax)

Cos²(ax) = Cos(ax).Cos(ax)
     = ½(Cos(ax+ax) + Cos(ax-ax))
     = ½(Cos(2ax) + Cos(0))
     = ½(Cos(2ax) + 1)
Cos²(ax) = ½Cos(2ax) + ½

(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x²; dv = ½Cos(2ax) + ½; du = 2x.dx; v = ¼Sin(2ax)/a + ½x
∫x².Cos²(ax) = x².(¼.Sin(2ax)/a + ½x) – ∫(¼Sin(2ax)/a + ½x) . 2x.dx
     = ¼.x².Sin(2ax)/a + ½x³ – ∫(½.x.Sin(2ax)/a + x²).dx
     = ¼.x².Sin(2ax)/a + ½x³ – ∫½.x.Sin(2ax)/a.dx – ∫x².dx
     = ¼.x².Sin(2ax)/a + ½x³ – ∫x².dx – 1 / 2a∫x.Sin(2ax).dx
     = ¼.x².Sin(2ax)/a + ½x³ – ⅓x³ – 1 / 2a∫x.Sin(2ax).dx
∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 – 1 / 2a∫x.Sin(2ax).dx

∫x.Sin(2ax).dx
u = x; dv = Sin(2ax); du = dx; v = -Cos(2ax) / 2a
∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a - ∫-Cos(2ax) / 2a . dx
     = -x.Cos(2ax) / 2a + 1 / 2a∫Cos(2ax) . dx
     = -x.Cos(2ax) / 2a + 1 / 2a.Sin(2ax) / 2a . dx
∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a + Sin(2ax) / 4a²

∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 – 1 / 2a . (-x.Cos(2ax) / 2a + Sin(2ax) / 4a²)
     = ¼.x².Sin(2ax)/a + x³/6 – (-x.Cos(2ax) / 4a² + Sin(2ax) / 8a³)
∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 + x.Cos(2ax) / 4a² – ⅛Sin(2ax)/a³

∫x³.Cos(ax)
(using integration by parts: ∫u.dv = uv - ∫v.du)
u = x³; dv = Cos(ax); du =3x².dx; v = Sin(ax)/a
∫x³.Cos(ax) = x³.Sin(ax)/a – ∫Sin(ax)/a . 3x².dx
∫x³.Cos(ax) = x³.Sin(ax)/a – 3/a∫x².Sin(ax).dx

∫x².Sin(ax).dx
u = x²; dv = Sin(ax); du =2x.dx; v = -Cos(ax)/a
∫x².Sin(ax).dx = x².-Cos(ax)/a – ∫-Cos(ax)/a . 2x.dx
     = -x².Cos(ax)/a + 2/a∫Cos(ax) . x.dx
∫x².Sin(ax).dx = -x².Cos(ax)/a + 2/a∫x.Cos(ax).dx

∫x.Cos(ax).dx
u = x; dv = Cos(ax); du =dx; v = Sin(ax)/a
∫x.Cos(ax).dx = x . Sin(ax)/a – ∫Sin(ax)/a . dx
     = x . Sin(ax)/a – 1/a∫Sin(ax).dx
     = x.Sin(ax)/a + Cos(ax)/a/a
∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a²

∫x².Sin(ax).dx = -x².Cos(ax)/a + 2/a . (x.Sin(ax)/a + Cos(ax)/a²)
     = -x².Cos(ax)/a + (2/a . x.Sin(ax)/a + 2/a . Cos(ax)/a²)
     = -x².Cos(ax)/a + (2.x.Sin(ax)/a² + 2.Cos(ax)/a³)
∫x².Sin(ax).dx = -x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2.Cos(ax)/a³

∫x³.Cos(ax) = x³.Sin(ax)/a – 3/a . (-x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2.Cos(ax)/a³)
     = x³.Sin(ax)/a – (3/a . -x².Cos(ax)/a + 3/a . 2.x.Sin(ax)/a² + 3/a . 2.Cos(ax)/a³)
     = x³.Sin(ax)/a – (-3x².Cos(ax)/a² + 6.x.Sin(ax)/a³ + 6.Cos(ax)/a⁴)
∫x³.Cos(ax) = x³.Sin(ax)/a + 3x².Cos(ax)/a² – 6.x.Sin(ax)/a³ – 6.Cos(ax)/a⁴

∫Sin(x).Cos(x).dx

Sin(x).Cos(x) = ½(Sin(x+x) + Sin(x-x))
     = ½(Sin(2x) + Sin(0))
     = ½(Sin(2x) + 0)
     = ½Sin(2x)

∫Sin(x).Cos(x) = ∫½Sin(2x).dx
     = ½∫Sin(2x).dx
     = ½.-Cos(2x)/2
∫Sin(x).Cos(x) = -¼.Cos(2x)

∫Tan²(x).dx

Tan²(x) = Sec²(x) – 1

∫Tan²(x) = ∫(Sec²(x) – 1).dx
     = ∫Sec²(x).dx – ∫dx
∫Tan²(x) = Tan(x) – x

Colour Coding is provided in the above table to assist with the flow/sequencing of some of the more complex calculations.

Further Reading

You will find further reading on this subject in reference publications(19)

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