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:
∫[a, b] f(x) dx = ∫[a, b] f(t) dt∫[a, b] f(x) dx = - ∫[b, a] f(x) dx∫[a, a] f(x) dx = 0∫[a, b] f(x) dx = ∫[a, c] f(x) dx + ∫[c, b] f(x) dx (for any c)∫[a, b] f(x) dx = ∫[a, b] f(a + b - x) dx
∫[0, a] f(x) dx = ∫[0, a] f(a - x) dx[-a, a]:
f(-x) = f(x)): ∫[-a, a] f(x) dx = 2 · ∫[0, a] f(x) dxf(-x) = -f(x)): ∫[-a, a] f(x) dx = 0This theorem provides the powerful connection between differentiation and integration, giving us the method to evaluate definite integrals.
Statement (Part 2): Iff(x)is a continuous function on[a, b]andF(x)is any antiderivative off(x)(meaningF'(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.
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.
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
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
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
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).
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.
This is a critical shortcut for evaluating definite integrals of sin and cos from 0 to π/2, derived from the reduction formulae.
((n-1)/n) · ((n-3)/(n-2)) · ... · (4/5) · (2/3)((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
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
sinⁿ, cosⁿ, secⁿ, (log x)ⁿ, Integration by Parts is the main method for deriving the reduction formulae.tanⁿ(x), split as tanⁿ⁻²(x)·tan²(x) and use tan²(x) = sec²(x) - 1. This is much faster than IBP.π/2. Remember it only applies to integrals from 0 to π/2.∫ sinᵐ(x) cosⁿ(x) dx from 0 to π/2, remember the π/2 factor only appears if BOTH m AND n are even.∫[0, a] f(x) dx = ∫[0, a] f(a - x) dx.