Reduction formulae for the evaluation of definite integrals

The idea. Suppose you must evaluate $I_5 = \int_0^1 x^5 e^x\,dx$. Integrating $x^5 e^x$ directly needs integration by parts five times — slow and error-prone. A reduction formula does the parts step once, in general, producing a relation like

$I_n = e - n\,I_{n-1}$

where $I_n = \int_0^1 x^n e^x\,dx$. Now you compute $I_0$ once and climb to $I_5$ by repeated substitution. The same single derivation handles every power.

The structure of a reduction formula:

• $I_n$ — the integral you want, depending on an integer index $n$.
• A relation expressing $I_n$ in terms of $I_{n-1}$ (drops by one) or $I_{n-2}$ (drops by two — typical for $\sin^n$, $\cos^n$, $\tan^n$).
• A base case — $I_0$ or $I_1$ (and sometimes both, when the step is $n \to n-2$) — that you evaluate directly.

Cambridge tip. Always write down the base case and check the index step (does $n$ drop by 1 or by 2?). For $\sin^n$ the step is $n \to n-2$, so an even $n$ bottoms out at $I_0$ and an odd $n$ at $I_1$.

Almost every reduction formula in 9231 is derived by integration by parts:

$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$

The standard move. Choose $u$ to be the part whose power you want to lower, so that differentiating it reduces the index. The rest becomes $\frac{dv}{dx}$.

Worked derivation — $I_n = \int x^n e^x\,dx$. Take $u = x^n$ (so $\frac{du}{dx} = nx^{n-1}$, index drops) and $\frac{dv}{dx} = e^x$ (so $v = e^x$):

$I_n = x^n e^x - \int nx^{n-1}e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx$

The remaining integral is exactly $I_{n-1}$, so

$I_n = x^n e^x - n\,I_{n-1}$

For the definite version $\int_0^1 x^n e^x\,dx$, apply the limits to the $x^n e^x$ term: $[x^n e^x]_0^1 = e - 0 = e$, giving $I_n = e - n\,I_{n-1}$.

Trickier cases need a Pythagorean split. For $I_n = \int \sin^n x\,dx$, write $\sin^n x = \sin^{n-1}x\cdot\sin x$, take $u = \sin^{n-1}x$, $\frac{dv}{dx} = \sin x$. After parts you get a $\cos^2 x$ term; replace $\cos^2 x = 1 - \sin^2 x$ and the integral of $\sin^n x$ (i.e. $I_n$) reappears on the right. Collect it:

$I_n = \frac{n-1}{n}\,I_{n-2}$ (for $\int_0^{\pi/2}\sin^n x\,dx$, the boundary term vanishes).

Cambridge tip. In a "Show that" derivation, state your choice of $u$ and $\frac{dv}{dx}$ explicitly, then show every line. The most-credited move — replacing $\cos^2 x$ by $1 - \sin^2 x$ so $I_n$ reappears — must be visible.

Once you have the formula, the evaluation is mechanical — but organise it carefully to avoid arithmetic slips.

The procedure:

1. Identify the base case ($I_0$ or $I_1$) and evaluate it directly.
2. Apply the reduction formula repeatedly, substituting numbers as you go.
3. Combine to reach the required $I_n$.

Worked application — $I_n = e - n\,I_{n-1}$, find $I_3 = \int_0^1 x^3 e^x\,dx$.

Base case: $I_0 = \int_0^1 e^x\,dx = [e^x]_0^1 = e - 1$.

Climb: $I_1 = e - 1\cdot I_0 = e - (e - 1) = 1$ $I_2 = e - 2 I_1 = e - 2$ $I_3 = e - 3 I_2 = e - 3(e - 2) = e - 3e + 6 = 6 - 2e$

So $I_3 = 6 - 2e \approx 0.563$ (3 s.f.).

When the step is $n \to n-2$ (e.g. $\int_0^{\pi/2}\sin^n x\,dx$, $I_n = \frac{n-1}{n}I_{n-2}$): the parity of $n$ decides the base case. To find $I_6$, use $I_0 = \frac{\pi}{2}$ and chain $I_6 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}$ — this is Wallis's formula. For odd $n$, stop at $I_1 = 1$.

Cambridge tip. Lay the recursion out as a vertical chain, one line per step, substituting the previous numerical value immediately. Carrying a symbolic $I_{n-1}$ too far is where sign and arithmetic errors creep in.

Some exam integrals carry two indices, e.g. $I_{n} = \int_0^1 x^n(1-x)^m\,dx$ for a fixed $m$. The reduction still works by parts — you lower one index at a time.

Worked shape. Take $u = (1-x)^m$, $\frac{dv}{dx} = x^n$, so $v = \frac{x^{n+1}}{n+1}$ and $\frac{du}{dx} = -m(1-x)^{m-1}$:

$I = \left[\frac{x^{n+1}}{n+1}(1-x)^m\right]_0^1 + \frac{m}{n+1}\int_0^1 x^{n+1}(1-x)^{m-1}\,dx$

The boundary term vanishes at both ends (at $x=0$ the $x^{n+1}$ kills it; at $x=1$ the $(1-x)^m$ kills it), leaving a clean relation that lowers $m$ by one while raising $n$.

Why this matters. Recognising that the boundary term is zero is itself a marked step. These integrals connect to the Beta function, $\int_0^1 x^{n}(1-x)^{m}\,dx = \frac{n!\,m!}{(n+m+1)!}$, though Cambridge expects you to derive and use the recursion, not quote that closed form.

General discipline for any reduction question:

• Choose $u$ to drop the index you're asked to reduce.
• Always check the boundary term $[uv]$ — often (but not always) zero for definite integrals over symmetric or vanishing endpoints.
• Show that means show that: state $u$, $\frac{dv}{dx}$, the parts line, the identity substitution, and the collection of $I_n$.

Cambridge tip. When a boundary term vanishes, say so explicitly — "the $[uv]$ term is $0$ at both limits" — rather than silently dropping it. Examiners award that observation.