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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