Unit 2: Reduction Formulae

Introduction to Reduction Formulae

A reduction formula is a recursive formula that expresses an integral Iₙ (involving a parameter n) in terms of a similar integral with a lower parameter, like Iₙ₋₁ or Iₙ₋₂.

The primary tool used to derive most reduction formulae is Integration by Parts (IBP):

∫ u dv = uv - ∫ v du

Derivation for Iₙ = ∫ sinⁿ(x) dx

1. Split the integrand: Iₙ = ∫ sinⁿ⁻¹(x) · sin(x) dx
2. Apply IBP:
   • Let u = sinⁿ⁻¹(x) → du = (n-1)sinⁿ⁻²(x)·cos(x) dx
   • Let dv = sin(x) dx → v = -cos(x)
3. Iₙ = uv - ∫ v du
Iₙ = -sinⁿ⁻¹(x)cos(x) - ∫ -cos(x) · (n-1)sinⁿ⁻²(x)cos(x) dx
4. Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ sinⁿ⁻²(x)cos²(x) dx
5. Use identity cos²(x) = 1 - sin²(x):
   Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ sinⁿ⁻²(x)(1 - sin²(x)) dx
6. Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) [∫ sinⁿ⁻²(x) dx - ∫ sinⁿ(x) dx]
7. Replace integrals with Iₙ₋₂ and Iₙ:
   Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂ - (n-1)Iₙ
8. Solve for Iₙ:
   Iₙ + (n-1)Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂
n·Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂

Iₙ = - (sinⁿ⁻¹(x) cos(x)) / n + (n-1)/n · Iₙ₋₂

Derivation for Iₙ = ∫ cosⁿ(x) dx

The derivation is perfectly analogous to sinⁿ(x). Split into cosⁿ⁻¹(x)·cos(x) and use IBP.

Iₙ = (cosⁿ⁻¹(x) sin(x)) / n + (n-1)/n · Iₙ₋₂

Derivation for Iₙ = ∫ tanⁿ(x) dx

This derivation uses a trick, not IBP.

1. Split the integrand: Iₙ = ∫ tanⁿ⁻²(x) · tan²(x) dx
2. Use identity tan²(x) = sec²(x) - 1:
   Iₙ = ∫ tanⁿ⁻²(x) · (sec²(x) - 1) dx
3. Distribute: Iₙ = ∫ tanⁿ⁻²(x)sec²(x) dx - ∫ tanⁿ⁻²(x) dx
4. The second integral is Iₙ₋₂.
5. The first integral is a standard u-substitution: let u = tan(x), so du = sec²(x) dx. The integral becomes ∫ uⁿ⁻² du = uⁿ⁻¹ / (n-1) = tanⁿ⁻¹(x) / (n-1).

Iₙ = tanⁿ⁻¹(x) / (n-1) - Iₙ₋₂

Derivation for Iₙ = ∫ secⁿ(x) dx

This derivation uses IBP.

1. Split the integrand: Iₙ = ∫ secⁿ⁻²(x) · sec²(x) dx
2. Apply IBP:
   • Let u = secⁿ⁻²(x) → du = (n-2)secⁿ⁻³(x)·sec(x)tan(x) dx = (n-2)secⁿ⁻²(x)tan(x) dx
   • Let dv = sec²(x) dx → v = tan(x)
3. Iₙ = secⁿ⁻²(x)tan(x) - ∫ tan(x) · (n-2)secⁿ⁻²(x)tan(x) dx
4. Iₙ = secⁿ⁻²(x)tan(x) - (n-2) ∫ secⁿ⁻²(x)tan²(x) dx
5. Use identity tan²(x) = sec²(x) - 1:
   Iₙ = secⁿ⁻²(x)tan(x) - (n-2) ∫ secⁿ⁻²(x)(sec²(x) - 1) dx
6. Iₙ = secⁿ⁻²(x)tan(x) - (n-2) [∫ secⁿ(x) dx - ∫ secⁿ⁻²(x) dx]
7. Iₙ = secⁿ⁻²(x)tan(x) - (n-2)Iₙ + (n-2)Iₙ₋₂
8. Solve for Iₙ:
   Iₙ + (n-2)Iₙ = secⁿ⁻²(x)tan(x) + (n-2)Iₙ₋₂
(n-1)Iₙ = secⁿ⁻²(x)tan(x) + (n-2)Iₙ₋₂

Iₙ = (secⁿ⁻²(x) tan(x)) / (n-1) + (n-2)/(n-1) · Iₙ₋₂

Derivation for Iₙ = ∫ (log x)ⁿ dx

This is a straightforward IBP.

1. Split the integrand: Iₙ = ∫ (log x)ⁿ · 1 dx
2. Apply IBP:
   • Let u = (log x)ⁿ → du = n(log x)ⁿ⁻¹ · (1/x) dx
   • Let dv = 1 dx → v = x
3. Iₙ = uv - ∫ v du
Iₙ = x(log x)ⁿ - ∫ x · n(log x)ⁿ⁻¹ (1/x) dx
4. Iₙ = x(log x)ⁿ - n ∫ (log x)ⁿ⁻¹ dx

Iₙ = x(log x)ⁿ - n · Iₙ₋₁

Derivation for Iₘ,ₙ = ∫ sinᵐ(x) cosⁿ(x) dx

This is the most complex reduction, with multiple possible formulas connecting Iₘ,ₙ to Iₘ₋₂,ₙ, Iₘ,ₙ₋₂, etc. We will derive one (connecting to Iₘ₋₂,ₙ).

1. Apply IBP:
   • Let u = sinᵐ⁻¹(x)
   • Let dv = sin(x)cosⁿ(x) dx → v = -cosⁿ⁺¹(x) / (n+1)
2. Iₘ,ₙ = - (sinᵐ⁻¹(x)cosⁿ⁺¹(x)) / (n+1) - ∫ - (cosⁿ⁺¹(x))/(n+1) · (m-1)sinᵐ⁻²(x)cos(x) dx
3. Iₘ,ₙ = - (sinᵐ⁻¹(x)cosⁿ⁺¹(x)) / (n+1) + (m-1)/(n+1) ∫ sinᵐ⁻²(x)cosⁿ⁺²(x) dx
4. Use identity cos²(x) = 1 - sin²(x):
   ... + (m-1)/(n+1) ∫ sinᵐ⁻²(x)cosⁿ(x)(1 - sin²(x)) dx
5. ... + (m-1)/(n+1) [∫ sinᵐ⁻²(x)cosⁿ(x) dx - ∫ sinᵐ(x)cosⁿ(x) dx]
6. Iₘ,ₙ = - (sinᵐ⁻¹(x)cosⁿ⁺¹(x)) / (n+1) + (m-1)/(n+1) [Iₘ₋₂,ₙ - Iₘ,ₙ]
7. Solve for Iₘ,ₙ (This is complex. A more common derivation links all terms).

A more standard and useful relation is found by letting u = sinᵐ⁺¹(x)cosⁿ⁺¹(x) and differentiating. The most common reduction formulas are:

Iₘ,ₙ = - (sinᵐ⁻¹(x) cosⁿ⁺¹(x)) / (m+n) + (m-1)/(m+n) · Iₘ₋₂,ₙ (connects m)

Iₘ,ₙ = (sinᵐ⁺¹(x) cosⁿ⁻¹(x)) / (m+n) + (n-1)/(m+n) · Iₘ,ₙ₋₂ (connects n)

Derivation for Iₘ,ₙ = ∫ sinᵐ(x) cos(nx) dx

This is a known, though advanced, reduction formula derived using IBP twice.

Applying IBP with u = sinᵐ(x) and dv = cos(nx) dx gives:
Iₘ,ₙ = (sinᵐ(x)sin(nx))/n - (m/n) ∫ sinᵐ⁻¹(x)cos(x)sin(nx) dx

Applying IBP *again* to the new integral (with u = sinᵐ⁻¹(x)cos(x)) and rearranging all terms eventually leads to a formula connecting Iₘ,ₙ with Iₘ₋₂,ₙ.

(n² - m²)Iₘ,ₙ = -m sinᵐ⁻¹(x)[m sin(x)cos(nx) + n cos(x)sin(nx)] + m(m-1)Iₘ₋₂,ₙ

This is a very complex reduction and is typically only studied in advanced calculus.

Wallis' Formula (for definite integrals)

A major application of reduction formulae for sinⁿ(x) and cosⁿ(x) is evaluating definite integrals from 0 to π/2.

∫ [from 0 to π/2] sinⁿ(x) dx = ∫ [from 0 to π/2] cosⁿ(x) dx

• If n is ODD (n = 3, 5, 7, ...):

  ((n-1)/n) · ((n-3)/(n-2)) · ... · (4/5) · (2/3)

• If n is EVEN (n = 2, 4, 6, ...):

  ((n-1)/n) · ((n-3)/(n-2)) · ... · (3/4) · (1/2) · (π/2)

Example (n=5, odd): ∫[0, π/2] sin⁵(x) dx = (4/5)·(2/3) = 8/15
Example (n=6, even): ∫[0, π/2] cos⁶(x) dx = (5/6)·(3/4)·(1/2)·(π/2) = 5π/32

∫ [from 0 to π/2] sinᵐ(x) cosⁿ(x) dx

Let m!! (double factorial) be m·(m-2)·(m-4)...

Iₘ,ₙ = ((m-1)!! · (n-1)!!) / ((m+n)!!) · k

Where:

• k = π/2 if both m and n are EVEN.
• k = 1 in all other cases (if m is odd, or n is odd, or both are odd).

Example (m=4, n=6, both even):
I₄,₆ = [(3·1)(5·3·1)] / [10·8·6·4·2] · (π/2) = (15 / 3840) · (π/2) = 3π/512

Example (m=5, n=3, both odd):
I₅,₃ = [(4·2)(2)] / [8·6·4·2] · 1 = 16 / 384 = 1/24

Unit 2: Exam Quick Tips