Integration by parts and reduction formulas: VCE Specialist Mathematics Units 3 & 4

1. Reverse the product rule

The product rule says (uv)′ = u′v + uv′. Integrating and rearranging isolates one product integral in terms of another: ∫u dv = uv − ∫v du. Integration by parts is therefore an exact rearrangement of a derivative identity. It does not say that the integral of a product is the product of two integrals.

Choosing u determines what you differentiate; choosing dv determines what you integrate to find v. The new integral should be simpler, or should return to a previous integral in a useful algebraic cycle. Always write u, du, dv and v explicitly at first. This small record prevents the common mistake of differentiating both factors or forgetting which expression supplies v.

For an indefinite calculation, choose any convenient antiderivative v without adding a separate constant at every intermediate line. Put one arbitrary C in the final family. Different choices of a constant in v cancel within the parts identity, but displaying them repeatedly obscures the calculation. For a definite integral there is no final C; the boundary product and the remaining definite integral use the same bounds.

2. Choose the part that becomes simpler

A polynomial usually becomes simpler when differentiated, while an exponential or sine/cosine remains easy to integrate. This makes u = polynomial a useful starting choice for products such as x²e²ˣ or x sin x. Differentiating the polynomial repeatedly eventually reaches a constant and then zero. The method ends after finitely many steps.

A logarithm becomes a rational function when differentiated. To integrate ln x, regard it as ln x × 1 and take u = ln x, dv = dx. Then du = dx/x and v = x, giving x ln x − ∫1 dx = x ln x − x + C for x > 0. The hidden factor 1 is legitimate and often unlocks an apparently single-function integral.

A mnemonic can suggest a choice but cannot prove it will help. Test the remaining integral before doing long algebra. If ∫xeˣ dx is assigned u = eˣ and dv = x dx, the new integral contains x²eˣ, increasing the polynomial degree. The identity remains true, but it moves in an unhelpful direction. Good method choice is about reducing the work, not satisfying a slogan.

Check whether substitution is simpler first. In ∫x e^x² dx, the derivative of x² is already present up to a constant, so substitution is direct. By contrast, ∫x eˣ dx has no matching inner derivative that removes the product; parts is natural. Reading the structure before choosing a technique is a central integration skill.

3. Definite parts includes a boundary term

When the question asks for a definite integral, working with the definite identity keeps the argument organised. Calculate [uv]_a^b carefully and then subtract the remaining integral over [a, b]. Do not evaluate just the new integral: the boundary product carries essential information. Bracket the entire lower-bound value before subtraction.

For example, ∫₁^e ln x dx = [x ln x]₁^e − ∫₁^e 1 dx = e − (e − 1) = 1. This offers a useful check on the indefinite formula above. Since ln x lies between zero and one on this interval, a positive area of one is plausible.

In applications, parts may convert an accumulated product into a boundary contribution minus another accumulation. This is an algebraic advantage, not a reason to omit interpretation. If the integrand is a rate in units per second, the final integral still has units, regardless of how many parts steps were used. Exact constants should remain exact until the requested rounding stage.

4. Build a reduction relation, rather than memorising one

A reduction relation connects an indexed integral to one with a smaller index. It is useful when repeating a method produces a predictable pattern. Define the family precisely, including its bounds and allowed integer values. Then derive the relation, state the base values and use it until a known base is reached. This connects integration with the recursive and proof ideas used elsewhere in Specialist Mathematics.

Let Iₙ = ∫₀^π/2 sinⁿx dx for integers n ≥ 0. For n ≥ 2, split sinⁿx into sinⁿ⁻¹x × sin x. Choose u = sinⁿ⁻¹x and dv = sin x dx. Then du = (n − 1)sinⁿ⁻²x cos x dx and v = −cos x. Both boundary products vanish: sin 0 = 0 and cos(π/2) = 0.

The substitution cos²x = 1 − sin²x exposes the earlier integral Iₙ again. Rearranging gives the reduction. The base values are I₀ = π/2 and I₁ = 1. Even indices descend to I₀, odd indices to I₁. The condition n ≥ 2 is part of the argument, because the derivation and the boundary reasoning were written for those powers.

5. Verification and common traps

The minus sign in uv − ∫v du is the most frequent source of trouble. Write the remaining integral in brackets if it is later replaced by several terms. Keep reverse chain-rule coefficients visible when v contains sin(ax), cos(ax) or eᵃˣ. A missing coefficient early in repeated parts contaminates every subsequent term.

For a symbolic answer, differentiate the whole expression and simplify before comparing with the integrand. For a recurrence, check one or two small admissible indices by a separate identity or direct integration. Numerical quadrature can verify a definite value to high precision, but it does not establish the recurrence for every index. The derivation supplies that general reasoning.

A short algorithm can implement reduction by starting with the correct parity base, increasing the index by two and multiplying by the corresponding factor. Preserve an exact rational multiplier if an exact answer is required. The stopping condition must be a known base or the target index, rather than an arbitrary number of iterations. Explain what the stored value represents at each stage.

6. Explain why a repeated method terminates

Repeated parts with a polynomial works because each differentiation lowers its non-negative integer degree. Starting from degree n, after n differentiations the polynomial becomes constant; one more derivative is zero. This supplies a mathematical stopping reason. It is stronger than observing that two steps happened to work in a particular example.

A cyclic integral has a different stopping reason. Neither exponential nor circular derivatives become zero, but a later step returns to the original integral multiplied by a known constant. Move that term to the other side and solve the resulting algebraic equation. The decision to stop comes from recognising a repeated object, not from waiting for the integrand to vanish.

A reduction relation combines these ideas with indexing. Its statement must identify which index decreases, how far it decreases and which base values are needed. A relation lowering n by two needs both an even and an odd starting case if both parities are requested. A correct relation with a missing base case does not yet evaluate the family.

In written solutions, distinguish deriving a relation from applying it. The derivation needs the parts choices, boundary reasoning and rearrangement. Application needs the appropriate index values and arithmetic. Technology can check either stage, but showing only a numerical value leaves the underlying pattern unexplained.