Reduction formulas are a standard tool in the integral calculus module of VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). Integrals such as ∫ sinⁿ x dx for large n are long to compute by hand, but each one can be reduced to an integral with a smaller power, then reduced again, until it becomes a basic integral. This guide derives the reduction formula for sinⁿ x, applies it to definite integrals, and shows how students should structure the working.
The Idea of a Reduction Formula
A reduction formula expresses an integral with power n in terms of the same kind of integral with a smaller power. Once you have the formula, you apply it repeatedly. For example, ∫ sin⁵ x dx reduces to ∫ sin³ x dx, which reduces to ∫ sin x dx, which is a basic integral. The method is systematic, and it is much more reliable than trying to remember the final answer for each power.
Deriving the Reduction Formula for sinⁿ x
Let Iₙ = ∫ sinⁿ x dx. Write sinⁿ x = sinⁿ⁻¹ x · sin x and integrate by parts with u = sinⁿ⁻¹ x and dv = sin x dx. Then du = (n−1) sinⁿ⁻² x cos x dx and v = −cos x.
Applying integration by parts:
Iₙ = −sinⁿ⁻¹ x cos x + (n−1) ∫ sinⁿ⁻² x cos² x dx
Replace cos² x by 1 − sin² x:
Iₙ = −sinⁿ⁻¹ x cos x + (n−1) ∫ sinⁿ⁻² x dx − (n−1) ∫ sinⁿ x dx
The last term is (n−1)Iₙ. Move it to the left side:
n Iₙ = −sinⁿ⁻¹ x cos x + (n−1) Iₙ₋₂
Dividing by n gives the standard reduction formula:
∫ sinⁿ x dx = −(sinⁿ⁻¹ x cos x)/n + ((n−1)/n) ∫ sinⁿ⁻² x dx
Students should know this formula well enough to rederive it quickly, because exam questions often ask for the derivation, not just the result.
The Definite Integral Form
For definite integrals over [0, π/2], the boundary term vanishes. At x = π/2, sin x = 1 and cos x = 0, so the term −sinⁿ⁻¹ x cos x equals zero. At x = 0, cos x = 1 but sin x = 0, so it is also zero for n ≥ 2. The formula simplifies to:
Iₙ = ∫₀^(π/2) sinⁿ x dx = ((n−1)/n) Iₙ₋₂
The starting values are I₀ = π/2 and I₁ = 1. Every other value follows by reduction. The result is known as Wallis’s formula, and it is a useful check on your answers.
Worked Example 1: ∫₀^(π/2) sin⁴ x dx
Apply the definite form with n = 4:
I₄ = (3/4) I₂
Now I₂ = (1/2) I₀ = (1/2)(π/2) = π/4.
So I₄ = (3/4)(π/4) = 3π/16.
Check this value using the general Wallis pattern for even powers: I₄ = (3·1)/(4·2) · π/2 = 3π/16. The results agree.
Worked Example 2: ∫₀^(π/2) sin⁵ x dx
For odd powers, the reduction ends at I₁ = 1.
I₅ = (4/5) I₃
I₃ = (2/3) I₁ = (2/3)(1) = 2/3.
So I₅ = (4/5)(2/3) = 8/15.
This is the pattern for odd powers: the result is a rational number, with no π, because the chain ends at I₁ = 1 rather than I₀ = π/2.
Worked Example 3: Indefinite Integral of sin³ x
Use the indefinite formula with n = 3:
∫ sin³ x dx = −(sin² x cos x)/3 + (2/3) ∫ sin x dx
= −(sin² x cos x)/3 − (2/3) cos x + C
Check by differentiating. d/dx[−(sin² x cos x)/3] = −(2 sin x cos² x − sin³ x)/3, and d/dx[−(2/3) cos x] = (2/3) sin x. Adding these gives (sin³ x − 2 sin x cos² x)/3 + (2/3) sin x. Replace cos² x by 1 − sin² x: (sin³ x − 2 sin x + 2 sin³ x)/3 + (2/3) sin x = sin³ x + (−2 sin x + 2 sin x)/3 = sin³ x. The derivative equals the integrand, so the answer is correct.
Worked Example 4: Reducing cos^n x
The same method applies to cosⁿ x. Over [0, π/2], the reduction formula is identical, because ∫₀^(π/2) cosⁿ x dx = ∫₀^(π/2) sinⁿ x dx by the substitution x → π/2 − x. So ∫₀^(π/2) cos⁶ x dx = ∫₀^(π/2) sin⁶ x dx = (5/6)(3/4)(1/2)(π/2) = 15π/96 = 5π/32.
Use this symmetry to save time in exams: any definite integral of cosⁿ x over [0, π/2] equals the corresponding sinⁿ x integral.
How to Set Up a Reduction Question
- Identify the power n. Note whether it is even or odd, because this determines the starting value.
- Decide whether the integral is definite or indefinite. Definite integrals over [0, π/2] use the simplified form.
- Reduce step by step. Write each step on its own line, and keep track of the coefficients.
- Substitute the base value. Use I₀ = π/2 for even n and I₁ = 1 for odd n.
- Check the result. Compare with the Wallis pattern for definite integrals, or differentiate for indefinite ones.
Common Mistakes That Cost Marks
Mistake 1: Using I₀ = 1 instead of π/2. I₀ is the integral of 1 over [0, π/2], which is π/2. Using 1 gives every even-power answer without π.
Mistake 2: Forgetting the coefficient (n−1)/n. Each reduction step multiplies by this fraction. Students who drop it get the wrong final answer, even if each step looks right.
Mistake 3: Dropping the boundary term in indefinite integrals. The term −sinⁿ⁻¹ x cos x is part of the answer. Omitting it is a frequent error in long questions.
Mistake 4: Stopping one step early. The reduction must continue until you reach I₀ or I₁. Stopping at I₂ or I₃ leaves the answer incomplete.
Practice Questions
- Evaluate ∫₀^(π/2) sin⁶ x dx. (Expected: 5π/32.)
- Evaluate ∫₀^(π/2) sin⁷ x dx. (Expected: 16/35.)
- Find ∫ sin⁴ x dx (indefinite). (Hint: apply the formula with n = 4 and reduce to ∫ sin² x dx, then use sin² x = (1 − cos 2x)/2.)
Checks: (1) I₆ = (5/6)(3/4)(1/2)(π/2) = 15π/96 = 5π/32 ✓. (2) I₇ = (6/7)(4/5)(2/3)(1) = 48/105 = 16/35 ✓. (3) The answer is (3x/8) − (sin 2x)/4 + (sin 4x)/32 + C. Differentiate to confirm.
Choosing Between a Reduction Formula and the Beta Function
Some integrals can be solved by either a reduction formula or the Beta function. Take I = ∫₀^(π/2) sin² x cos² x dx. Both routes give the same answer, and comparing them is good practice.
Route 1: Rewrite the integrand. Since sin² x cos² x = sin² x − sin⁴ x, the integral splits into two known integrals. From the reduction formula, ∫₀^(π/2) sin² x dx = π/4 and ∫₀^(π/2) sin⁴ x dx = 3π/16. So I = π/4 − 3π/16 = π/16.
Route 2: Beta function. Matching the exponents gives 2m − 1 = 2 and 2n − 1 = 2, so m = n = 3/2. Then I = (1/2) B(3/2, 3/2) = (1/2) Γ(3/2)² / Γ(3). Using Γ(3/2) = √π/2 and Γ(3) = 2, this is (1/2)(π/4)/2 = π/16.
Both routes agree. For integrals with a small integer power, the reduction route is usually faster. When the exponents are fractional, or when the integral is over [0, 1] in the form ∫x^(m−1)(1−x)^(n−1), the Beta function is the cleaner method. Knowing both lets you choose the faster route in an exam.
A Common Interval Check
Before applying the trigonometric Beta form, confirm the limits. The factor of 1/2 applies only over [0, π/2]. Over [0, π], the integral of sin^(2m−1) θ cos^(2n−1) θ takes a different form, and using the wrong factor changes the answer. Write the limits on your working before substituting, so that the correct form is chosen.
Connecting This Topic to the Full Module
Reduction formulas and the Beta and Gamma functions are both part of the integral calculus module of VTU Engineering Mathematics 1. The module outline, the standard integrals students should know, and the places where marks are typically lost are set out in our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Is the reduction formula the same for cos^n x?
The indefinite formula is different in form, but over [0, π/2] the definite integrals are equal, as shown above.
Q: Do I need to memorise the Wallis pattern?
Derive it from the reduction formula when needed. Knowing the pattern helps you check answers quickly in an exam.
Q: Where does this fit in the module?
It sits with the standard integrals and the Beta-Gamma functions. The full module outline is in our VTU Engineering Maths 1 notes guide.
For more solved integration problems, see the Engineering Math Survival Kit. To ask about batches covering this module, use the contact page.