One rule instead of ten substitutions
Imagine being asked for the antiderivative of sin^10(x). You could reach for trig identities and grind for an afternoon — or you could notice that this integral is not alone. It is one member of an infinite family indexed by an integer n: the integral of sin^n(x) dx, for n = 0, 1, 2, 3, and on up. The crucial fact is that the members are wired together. The integral with exponent n can be written in terms of the same integral with a smaller exponent. A [[reduction-formula|reduction formula]] is exactly that wire — a rule that lowers the power by one or two each time you apply it.
Write I_n for the integral of sin^n(x) dx. A reduction formula is a statement like I_n = (something simple) + (a fraction) times I_{n-2}. Notice what it does and does not do: it does not hand you the answer outright — it only shifts the index downward. But shift it enough times and you land on a base case, I_0 or I_1, that you already know cold. This self-referential style — defining a quantity in terms of smaller instances of itself — is the heart of [[recursive-integration|recursive integration]]. It is the same spirit as a recurrence relation in a sequence, only here the terms are whole integrals.
The engine room: integration by parts
Where does a reduction formula come from? Almost always from integration by parts, the tool you already met earlier in this rung. Recall its shape: the integral of u dv equals u v minus the integral of v du. The art is the split. To lower a power you peel off one factor to differentiate (that factor's exponent drops by one when you take its derivative) and leave the rest to integrate. Done cleverly, the leftover integral on the right is the original family member with a smaller exponent — and you have your recurrence.
Worked example: powers of sine
Take I_n = integral of sin^n(x) dx and split it as sin^(n-1)(x) times sin(x) dx. Let u = sin^(n-1)(x) and dv = sin(x) dx. Then du = (n-1) sin^(n-2)(x) cos(x) dx and v = -cos(x). Integration by parts gives I_n = -sin^(n-1)(x) cos(x) + (n-1) times the integral of sin^(n-2)(x) cos^2(x) dx. Now use the Pythagorean identity cos^2(x) = 1 - sin^2(x): that last integral becomes (n-1)(I_{n-2} - I_n). The original I_n has reappeared on the right — a recurring theme — so you collect it.
Gather the I_n terms: I_n + (n-1) I_n = -sin^(n-1)(x) cos(x) + (n-1) I_{n-2}, so n I_n = -sin^(n-1)(x) cos(x) + (n-1) I_{n-2}. Divide by n and you have the reduction formula: I_n = -(1/n) sin^(n-1)(x) cos(x) + ((n-1)/n) I_{n-2}. Each application drops the power by two. To get sin^4(x): I_4 uses I_2, which uses I_0 = integral of 1 dx = x. Three short lines and you are done — no afternoon of identities. This is the standard machine for any of the powers of trig functions.
There is a beautiful payoff if you make the integral a definite integral from 0 to pi/2. Then the boundary term sin^(n-1)(x) cos(x) vanishes at both ends, leaving the clean recurrence I_n = ((n-1)/n) I_{n-2}. Cascading this from a high n down to I_0 = pi/2 or I_1 = 1 produces the famous Wallis product for pi — a striking sign that a humble reduction formula can carry real structure, not just bookkeeping.
Worked example: powers of secant
Secant is trickier because there is no friendly base case to integrate at the very bottom, and the split is less obvious. For J_n = integral of sec^n(x) dx, peel off sec^2(x) as the part to integrate (it integrates to tan(x)) and let u = sec^(n-2)(x). Then du = (n-2) sec^(n-2)(x) tan(x) dx and dv = sec^2(x) dx, v = tan(x). Parts gives J_n = sec^(n-2)(x) tan(x) - (n-2) times the integral of sec^(n-2)(x) tan^2(x) dx.
Now use tan^2(x) = sec^2(x) - 1, so the trailing integral splits into J_n - J_{n-2}. As before, J_n returns on the right; collect it. You get (n-1) J_n = sec^(n-2)(x) tan(x) + (n-2) J_{n-2}, hence J_n = (1/(n-1)) sec^(n-2)(x) tan(x) + ((n-2)/(n-1)) J_{n-2}. The power again falls by two. The base cases sit one level up from sine's: J_2 = tan(x), and J_1 = integral of sec(x) dx = ln|sec(x) + tan(x)| — a small standard result worth memorizing, since the recursion for odd powers terminates there.
Two flavours of recursion: marching vs. looping
Now x times e^x — a mixed product. For K_n = integral of x^n e^x dx, the differentiable factor is x^n: choose u = x^n, dv = e^x dx, so du = n x^(n-1) dx and v = e^x. Parts gives K_n = x^n e^x - n K_{n-1}. This one is even simpler: the power drops by one each time, marching K_3 -> K_2 -> K_1 -> K_0 = e^x. For example K_1 = integral of x e^x dx = x e^x - K_0 = x e^x - e^x = (x - 1) e^x. The recursion terminates the moment the exponent on x hits zero.
The sine, secant, and x^n e^x cases all march downward to a base case. But recursive integration has a second flavour that never marches: the self-closing loop. The textbook case is the integral of e^x cos(x) dx. Apply parts once and you get an integral of e^x sin(x); apply parts again and the original e^x cos(x) integral reappears with a sign flip. Calling the original L, you reach an equation L = e^x cos(x) + e^x sin(x) - L. There is no base case here at all — you simply solve the algebra: 2L = e^x (cos x + sin x), so L = e^x (cos x + sin x) / 2.
I_n = -(1/n) sin^(n-1)(x) cos(x) + ((n-1)/n) I_{n-2} base: I_0 = x, I_1 = -cos(x)
J_n = (1/(n-1)) sec^(n-2)(x) tan(x) + ((n-2)/(n-1)) J_{n-2} base: J_1 = ln|sec x + tan x|, J_2 = tan x
K_n = x^n e^x - n K_{n-1} base: K_0 = e^x
L = integral e^x cos x dx -> L = e^x cos x + e^x sin x - L -> L = e^x(cos x + sin x)/2Pitfalls, base cases, and where this leads
- Name the family: write I_n for the integral and identify the integer index you want to lower.
- Split for integration by parts so the factor whose power you attack is the one you differentiate.
- Apply parts, then use an identity (like cos^2 = 1 - sin^2) to coax the original I_n back onto the right-hand side.
- Collect the I_n terms and solve algebraically for I_n in terms of I_{n-1} or I_{n-2}.
- Iterate down to the base case (I_0 or I_1), substitute it in, and back-fill upward.
This recursive habit reaches well past these examples. The definite sin^n integral over 0 to pi/2 is exactly the kind of object the gamma function and beta function package in closed form: B(p, q) = 2 times the integral from 0 to pi/2 of (sin theta)^(2p-1) (cos theta)^(2q-1) d theta, with B(p, q) = Gamma(p) Gamma(q) / Gamma(p+q). And the very logic of defining each term from earlier ones is the same recurrence relation you will lean on when you build power-series solutions to differential equations later in this volume. A reduction formula is your first real taste of that idea — recursion as a method, not just a trick.