Unit 4: Definite Integrals and Reduction Formulae

Definite Integrals

Definition and Properties

A definite integral, denoted ∫[a, b] f(x) dx, represents the signed area under the curve y = f(x) from x = a to x = b.

Key Properties:

1. Change of Variable: ∫[a, b] f(x) dx = ∫[a, b] f(t) dt
2. Reversing Limits: ∫[a, b] f(x) dx = - ∫[b, a] f(x) dx
3. Zero Width: ∫[a, a] f(x) dx = 0
4. Additivity: ∫[a, b] f(x) dx = ∫[a, c] f(x) dx + ∫[c, b] f(x) dx (for any c)
5. King's Property: ∫[a, b] f(x) dx = ∫[a, b] f(a + b - x) dx
   • Special Case (0 to a): ∫[0, a] f(x) dx = ∫[0, a] f(a - x) dx
6. Even/Odd Functions: For a symmetric interval [-a, a]:
   • If f(x) is Even (f(-x) = f(x)): ∫[-a, a] f(x) dx = 2 · ∫[0, a] f(x) dx
   • If f(x) is Odd (f(-x) = -f(x)): ∫[-a, a] f(x) dx = 0

Fundamental Theorem of Calculus (Statement)

This theorem provides the powerful connection between differentiation and integration, giving us the method to evaluate definite integrals.

Statement (Part 2): If f(x) is a continuous function on [a, b] and F(x) is any antiderivative of f(x) (meaning F'(x) = f(x)), then:
∫[a, b] f(x) dx = F(b) - F(a)

This is why we find the indefinite integral first, then plug in the upper and lower bounds.

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Reduction Formulae

Introduction

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 them is Integration by Parts (IBP): ∫ u dv = uv - ∫ v du.

∫ sinⁿ(x) dx and ∫ cosⁿ(x) dx

Let Iₙ = ∫ sinⁿ(x) dx. We use IBP by splitting sinⁿ(x) = sinⁿ⁻¹(x) · sin(x).

Reduction Formula for sinⁿ(x):
Iₙ = - (sinⁿ⁻¹(x) cos(x)) / n + (n-1)/n · Iₙ₋₂

Similarly, for Jₙ = ∫ cosⁿ(x) dx:

Reduction Formula for cosⁿ(x):
Jₙ = (cosⁿ⁻¹(x) sin(x)) / n + (n-1)/n · Jₙ₋₂

∫ tanⁿ(x) dx and ∫ secⁿ(x) dx

Let Iₙ = ∫ tanⁿ(x) dx. This derivation uses a trick, not IBP.

Write Iₙ = ∫ tanⁿ⁻²(x) · tan²(x) dx = ∫ tanⁿ⁻²(x) (sec²(x) - 1) dx

Reduction Formula for tanⁿ(x):
Iₙ = tanⁿ⁻¹(x) / (n-1) - Iₙ₋₂

Let Jₙ = ∫ secⁿ(x) dx. This uses IBP with secⁿ(x) = secⁿ⁻²(x) · sec²(x).

Reduction Formula for secⁿ(x):
Jₙ = (secⁿ⁻²(x) tan(x)) / (n-1) + (n-2)/(n-1) · Jₙ₋₂

∫ (log x)ⁿ dx

Let Iₙ = ∫ (log x)ⁿ dx. We use IBP with u = (log x)ⁿ and dv = 1 dx.

Reduction Formula for (log x)ⁿ:
Iₙ = x(log x)ⁿ - n · Iₙ₋₁

∫ sinᵐ(x) cosⁿ(x) dx

This is a more complex reduction with multiple possible forms. A common one connects Iₘ,ₙ to Iₘ,ₙ₋₂:

Reduction Formula for sinᵐ(x) cosⁿ(x):
Iₘ,ₙ = (sinᵐ⁺¹(x) cosⁿ⁻¹(x)) / (m+n) + (n-1)/(m+n) · Iₘ,ₙ₋₂

(Another formula connects to Iₘ₋₂,ₙ, and is also valid).

∫ sinᵐ(x) cos(nx) dx

This formula is derived using IBP twice and is more advanced.

Let Iₘ,ₙ = ∫ sinᵐ(x) cos(nx) dx

Reduction Formula:
Iₘ,ₙ = (sinᵐ(x)sin(nx))/n - (m/n) ∫ sinᵐ⁻¹(x)cos(x)sin(nx) dx

Further reduction of the new integral is required, leading to a formula connecting Iₘ,ₙ with Iₘ₋₂,ₙ.

Wallis' Formula (for definite integrals)

This is a critical shortcut for evaluating definite integrals of sin and cos from 0 to π/2, derived from the reduction formulae.

Case 1: ∫ [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

Case 2: ∫ [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

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Unit 4: Exam Quick Tips