reduction formula
Suppose you need the integral of sin^10(x), or x^6 e^x, or sec^5(x). Doing it from scratch is brutal, but each of these belongs to a family indexed by an integer n, and the members are linked: the integral with exponent n can be expressed using the same integral with a smaller exponent. A reduction formula is exactly that link — a rule that drops the exponent by one or two, so that repeated application walks you down to a base case you can do in your head.
You derive a reduction formula almost always by integration by parts, choosing the split so that the resulting integral is the original with a lowered power. For instance, calling I_n the integral of sin^n(x) dx, integration by parts yields I_n = -(1/n) sin^(n-1)(x) cos(x) + ((n-1)/n) I_{n-2}. Each use cuts the power by 2, marching toward I_0 = x or I_1 = -cos(x). The formula is a recurrence: a self-referential rule that defines a quantity in terms of smaller instances of itself.
Reduction formulas are how computer algebra systems and table-makers handle whole infinite families at once, and they make otherwise hopeless high-power integrals routine. They also reveal structure — for example the famous Wallis product for pi falls straight out of the sin^n reduction formula evaluated over 0 to pi/2. The catch is only that you must apply the formula faithfully down to the correct base case; an off-by-one in the base term silently corrupts the whole chain.
For I_n = integral of sin^n(x) dx, integration by parts gives I_n = -(1/n) sin^(n-1)(x) cos(x) + ((n-1)/n) I_{n-2}; applying it twice reduces sin^4 down to I_0 = x.
One integration-by-parts step lowers the power by two, so iterating it solves the whole family.
A reduction formula does not evaluate the integral by itself — it only shifts the index. You still need a correct base case (I_0 or I_1) at the bottom, or the recursion has nothing to stand on.