A definite integral represents the area under a curve f(x) from x = a to x = b. It is formally defined as the limit of a Riemann sum.
∫ [a to b] f(x) dx = lim (n→∞) Σ [i=1 to n] f(xᵢ*) * Δx Where:
- [a, b] is divided into n subintervals.
- Δx = (b - a) / n is the width of each subinterval.
- xᵢ* is a sample point in the i-th subinterval.
This theorem (stated without proof) provides the "fundamental" link between differentiation and integration.
If f is continuous on [a, b], then the function G(x) defined by G(x) = ∫ [a to x] f(t) dt is continuous on [a, b], differentiable on (a, b), and G'(x) = f(x).
This is the part we use for calculations. If f is continuous on [a, b] and F(x) is any antiderivative of f(x) (meaning F'(x) = f(x)), then:
∫ [a to b] f(x) dx = F(b) - F(a)
A reduction formula is an integration formula that expresses an integral involving a power of a function (say, n) in terms of an integral involving a lower power of that same function (e.g., n-1 or n-2).
The primary method to derive these formulas is Integration by Parts:
∫ u dv = uv - ∫ v du
The key is to choose 'u' and 'dv' correctly.
Let Iₙ = ∫ sinⁿ(x) dx
We can write this as Iₙ = ∫ sinⁿ⁻¹(x) * sin(x) dx
Using integration by parts:
Iₙ = [sinⁿ⁻¹(x)](-cos x) - ∫ (-cos x)[(n-1)sinⁿ⁻²(x)cos(x)] dx
Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ cos²(x)sinⁿ⁻²(x) dx
Replace cos²(x) with (1 - sin²(x)):
Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) ∫ (1 - sin²(x))sinⁿ⁻²(x) dx
Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1) [ ∫ sinⁿ⁻²(x) dx - ∫ sinⁿ(x) dx ]
Notice that the integrals are Iₙ₋₂ and Iₙ:
Iₙ = -sinⁿ⁻¹(x)cos(x) + (n-1)Iₙ₋₂ - (n-1)Iₙ
Now, 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
Reduction Formula: ∫ sinⁿ(x) dx = [-sinⁿ⁻¹(x)cos(x)] / n + [(n-1)/n] * ∫ sinⁿ⁻²(x) dx
Let Iₙ = ∫ cosⁿ(x) dx
The derivation is almost identical. We write Iₙ = ∫ cosⁿ⁻¹(x) * cos(x) dx
Reduction Formula: ∫ cosⁿ(x) dx = [cosⁿ⁻¹(x)sin(x)] / n + [(n-1)/n] * ∫ cosⁿ⁻²(x) dx
Let Iₙ = ∫ tanⁿ(x) dx. This one is different; we don't use integration by parts.
Write Iₙ = ∫ tanⁿ⁻²(x) * tan²(x) dx
Use the identity tan²(x) = sec²(x) - 1:
Iₙ = ∫ tanⁿ⁻²(x) * (sec²(x) - 1) dx
Iₙ = ∫ tanⁿ⁻²(x)sec²(x) dx - ∫ tanⁿ⁻²(x) dx
The second integral is just Iₙ₋₂.
For the first integral, let u = tan(x), so du = sec²(x) dx. The integral becomes ∫ uⁿ⁻² du = uⁿ⁻¹ / (n-1) = tanⁿ⁻¹(x) / (n-1).
Reduction Formula: ∫ tanⁿ(x) dx = [tanⁿ⁻¹(x)] / (n-1) - ∫ tanⁿ⁻²(x) dx
This is the most complex one. Let Iₙ,ₘ = ∫ sinⁿ(x)cosᵐ(x) dx.
There are several ways to derive this, connecting Iₙ,ₘ to Iₙ₋₂,ₘ or Iₙ,ₘ₋₂.
By choosing u = sinⁿ⁻¹(x) and dv = sin(x)cosᵐ(x) dx, we can derive:
Reduction Formula (reducing n): Iₙ,ₘ = [-sinⁿ⁻¹(x)cosᵐ⁺¹(x)] / (n+m) + [(n-1)/(n+m)] * Iₙ₋₂,ₘ
Alternatively, by choosing u = cosᵐ⁻¹(x) and dv = sinⁿ(x)cos(x) dx, we can derive:
Reduction Formula (reducing m): Iₙ,ₘ = [sinⁿ⁺¹(x)cosᵐ⁻¹(x)] / (n+m) + [(m-1)/(n+m)] * Iₙ,ₘ₋₂
Let Iₙ = ∫ (log x)ⁿ dx
We use integration by parts with dv = 1*dx.
Iₙ = x(log x)ⁿ - ∫ x * [n(log x)ⁿ⁻¹ * (1/x)] dx
Iₙ = x(log x)ⁿ - ∫ n(log x)ⁿ⁻¹ dx
Iₙ = x(log x)ⁿ - n * Iₙ₋₁
Reduction Formula: ∫ (log x)ⁿ dx = x(log x)ⁿ - n * ∫ (log x)ⁿ⁻¹ dx
This is a powerful shortcut for evaluating definite integrals of sin and cos from 0 to π/2.
It is derived by applying the reduction formulas for sinⁿ(x) and cosⁿ(x) to the definite integral ∫ [0 to π/2].
Let Iₙ = ∫ [0 to π/2] sinⁿ(x) dx = ∫ [0 to π/2] cosⁿ(x) dx
If n is EVEN (n = 2, 4, 6, ...): Iₙ = [ (n-1)/n ] * [ (n-3)/(n-2) ] * [ (n-5)/(n-4) ] * ... * [ 1/2 ] * (π/2)
If n is ODD (n = 3, 5, 7, ...): Iₙ = [ (n-1)/n ] * [ (n-3)/(n-2) ] * [ (n-5)/(n-4) ] * ... * [ 2/3 ] * (1)
Example (n=6, even):
∫ [0 to π/2] sin⁶(x) dx = (5/6) * (3/4) * (1/2) * (π/2) = 15π / 192 = 5π / 64
Example (n=5, odd):
∫ [0 to π/2] cos⁵(x) dx = (4/5) * (2/3) = 8 / 15
Let Iₙ,ₘ = ∫ [0 to π/2] sinⁿ(x)cosᵐ(x) dx
Iₙ,ₘ = [ (n-1)(n-3)...(1 or 2) ] * [ (m-1)(m-3)...(1 or 2) ] / [ (n+m)(n+m-2)...(1 or 2) ] * K
Where:
- K = π/2 if both n and m are EVEN.
- K = 1 otherwise (if at least one of them is odd).
Example (n=4, m=2; both even):
∫ [0 to π/2] sin⁴(x)cos²(x) dx = [ (3*1) * (1) ] / [ (6)(4)(2) ] * (π/2)
= [ 3 * 1 ] / [ 48 ] * (π/2) = 3π / 96 = π / 32
Example (n=5, m=3; both odd):
∫ [0 to π/2] sin⁵(x)cos³(x) dx = [ (4*2) * (2) ] / [ (8)(6)(4)(2) ] * (1)
= [ 8 * 2 ] / [ 384 ] = 16 / 384 = 1 / 24