Reduction Formulae

A reduction formula is a recurrence relation that expresses an integral $I_n$ — one carrying an integer parameter $n$ — in terms of a lower-index integral such as $I_{n-1}$ or $I_{n-2}$ . Rather than integrating $\int \sin^{8} x\,dx$ or $\int x^{6}e^{x}\,dx$ from scratch (a nightmare of repeated parts), you derive the recurrence once, then climb down to a base case you can evaluate instantly. This lesson builds the technique rigorously, drills the integration-by-parts derivations that earn the marks, and stretches towards the Wallis product and the Beta function.

1 Spec mapping (AQA 7367)

Reduction formulae sit in the compulsory Pure content assessed on Paper 1 and Paper 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), inside the "Further calculus" strand alongside improper integrals, volumes of revolution and the mean value of a function. The topic is unusually rich in AO2: a part-(a) "Show that $I_n = \ldots$ " is a proof, rewarding rigorous, well-communicated integration by parts, not just a final answer. The follow-up "Hence evaluate $I_4$ " is AO1 (apply your own recurrence accurately), and an unfamiliar integrand dressed up so you must spot that a reduction is the right tool reaches into AO3. Because the derivation is the marks, this is a topic where disciplined layout — stating $u$ , $\mathrm{d}v$ , $\mathrm{d}u$ , $v$ explicitly and handling the boundary term cleanly — converts directly into a full score.

2 Core theory — what a reduction formula is, and why it works

Let $I_n$ denote a definite or indefinite integral depending on a non-negative integer $n$ . A reduction (or recurrence) formula has the shape

$I_n = g(n) + h(n)\,I_{n-k}, \qquad k \in \{1,2\},$

where $g(n)$ is an explicit term (often a boundary contribution, sometimes $0$ ) and $h(n)$ is a coefficient in $n$ . Applying it repeatedly reduces the index until you reach a base case — $I_0$ or $I_1$ — that integrates trivially. The structure mirrors mathematical induction in reverse: where induction builds up from a base case, a reduction formula climbs down to one, and the recurrence is the inductive step that connects consecutive terms. The engine that produces almost every reduction formula is integration by parts,

$\int u\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\,\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x,$

with the parameter $n$ chosen to "peel off" one power. The art is splitting the integrand so that the leftover integral is a lower-index version of the same family. A handful of algebraic identities recur:

Identity	Used for
$\cos^2 x = 1-\sin^2 x$	$\int \sin^n x\,dx$
$\sin^2 x = 1-\cos^2 x$	$\int \cos^n x\,dx$
$\tan^2 x = \sec^2 x - 1$	$\int \tan^n x\,dx$
$\sec^2 x = 1+\tan^2 x$	$\int \sec^n x\,dx$

The standard workflow is: (1) write $I_n$ ; (2) integrate by parts, splitting off one factor; (3) use an identity to re-express the leftover integral in terms of $I_{n-1}$ or $I_{n-2}$ ; (4) rearrange into the recurrence; (5) find the base case(s); (6) climb down.

Why a recurrence, not a formula? For $\int\sin^{8}x\,\mathrm{d}x$ a direct attack would mean eight nested integrations by parts — error-prone and slow. The recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ does the same work once and for all, expressing the answer for any $n$ in terms of a smaller case. This is the same idea as a recurrence relation for a sequence: instead of a closed formula you have a rule that steps the index down, plus a starting value. The payoff is enormous — one derivation handles an entire infinite family of integrals.

The choice of which factor to differentiate and which to integrate matters. The guiding principle is the same LIATE-style logic as ordinary integration by parts: differentiate the part that simplifies (a power $x^n$ drops to $x^{n-1}$ ; a power $\sin^{n-1}x$ drops by one), and integrate the part that stays manageable ( $e^x$ stays $e^x$ ; $\sin x$ becomes $-\cos x$ ). Get this the wrong way round and the leftover integral becomes more complicated, not less, and no recurrence emerges. After the parts step, the leftover almost always needs one of the Pythagorean identities above to be massaged back into the same family — this re-expression is the second decisive move.

3 Worked examples with M1/A1 mark scheme

Worked Example 1 — $\displaystyle I_n = \int_0^{1} x^n e^{x}\,dx$

Take $u = x^n,\ \dfrac{\mathrm{d}v}{\mathrm{d}x} = e^{x}$ , so $\dfrac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1},\ v = e^{x}$ :

$I_n = \Big[ x^n e^{x} \Big]_0^{1} - \int_0^{1} n x^{n-1} e^{x}\,dx = \big( e - 0 \big) - n\int_0^{1} x^{n-1}e^{x}\,dx.$

(M1 for parts with the correct $u,v$ ; A1 for the boundary term $e$ and the lower integral.) The remaining integral is exactly $I_{n-1}$ , so

$\boxed{\,I_n = e - n\,I_{n-1}\,} \qquad (n \ge 1).$

Base case $I_0 = \displaystyle\int_0^{1} e^{x}\,dx = [e^{x}]_0^{1} = e-1$ . Climbing up:

$\begin{aligned} I_1 &= e - 1\cdot I_0 = e - (e-1) = 1, \\ I_2 &= e - 2 I_1 = e - 2, \\ I_3 &= e - 3 I_2 = e - 3(e-2) = 6 - 2e. \end{aligned}$

(M1 for correct iterative substitution; A1 for $I_2 = e-2$ and $I_3 = 6-2e$ .)

Worked Example 2 — the sine reduction $\displaystyle I_n = \int_0^{\pi/2} \sin^n x\,dx$

This is the single most important reduction formula in the course. Write $\sin^n x = \sin^{n-1}x\cdot \sin x$ and take $u = \sin^{n-1}x,\ \dfrac{\mathrm{d}v}{\mathrm{d}x} = \sin x$ ; then $\dfrac{\mathrm{d}u}{\mathrm{d}x} = (n-1)\sin^{n-2}x\cos x,\ v = -\cos x$ :

$I_n = \Big[ -\sin^{n-1}x\,\cos x \Big]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2}x\,\cos^2 x\,dx.$

(M1 for the split and parts; A1 for this line.) The boundary term vanishes at both ends (at $x=\tfrac{\pi}{2}$ because $\cos\tfrac{\pi}{2}=0$ ; at $x=0$ because $\sin 0 = 0$ ). Now substitute $\cos^2 x = 1-\sin^2 x$ :

$I_n = (n-1)\int_0^{\pi/2} \sin^{n-2}x\,(1-\sin^2 x)\,dx = (n-1)I_{n-2} - (n-1)I_n.$

(M1 for using the Pythagorean identity to recover $I_{n-2}$ and $I_n$ .) Collect the $I_n$ terms:

$I_n + (n-1)I_n = (n-1)I_{n-2} \;\implies\; n\,I_n = (n-1)I_{n-2} \;\implies\; \boxed{\,I_n = \frac{n-1}{n}\,I_{n-2}\,}.$

(A1 for the rearrangement to the printed form.) Base cases $I_0 = \int_0^{\pi/2} 1\,dx = \tfrac{\pi}{2}$ and $I_1 = \int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = 1$ . Then

$\begin{aligned} I_2 &= \tfrac12 I_0 = \tfrac{\pi}{4}, & I_3 &= \tfrac23 I_1 = \tfrac23, \\ I_4 &= \tfrac34 I_2 = \tfrac{3\pi}{16}, & I_5 &= \tfrac45 I_3 = \tfrac{8}{15}, \\ I_6 &= \tfrac56 I_4 = \tfrac{5\pi}{32}, & I_7 &= \tfrac67 I_5 = \tfrac{16}{35}. \end{aligned}$

Notice the pattern: even $n$ leaves a factor of $\pi$ ; odd $n$ is purely rational — because the descent ends on $I_0=\tfrac{\pi}{2}$ or $I_1=1$ respectively.

Worked Example 3 — the tangent reduction $\displaystyle I_n = \int_0^{\pi/4} \tan^n x\,dx$

Here the trick is an identity, not parts. Write $\tan^n x = \tan^{n-2}x\,(\sec^2 x - 1)$ :

$I_n = \int_0^{\pi/4} \tan^{n-2}x\,\sec^2 x\,dx - \int_0^{\pi/4} \tan^{n-2}x\,dx.$

(M1 for the $\sec^2 x - 1$ split.) The first integral yields to $u=\tan x$ ( $\mathrm{d}u = \sec^2 x\,dx$ ): $\Big[\tfrac{\tan^{n-1}x}{n-1}\Big]_0^{\pi/4} = \tfrac{1}{n-1}$ since $\tan\tfrac{\pi}{4}=1$ . The second is $I_{n-2}$ :

$\boxed{\,I_n = \frac{1}{n-1} - I_{n-2}\,} \qquad (n \ge 2).$

(A1 for the recurrence.) Base cases $I_0 = \tfrac{\pi}{4}$ and $I_1 = \int_0^{\pi/4}\tan x\,dx = [-\ln|\cos x|]_0^{\pi/4} = -\ln\tfrac{1}{\sqrt2} = \tfrac12\ln 2$ . Hence

$I_2 = 1 - \tfrac{\pi}{4}, \qquad I_3 = \tfrac12 - \tfrac12\ln 2, \qquad I_4 = \tfrac13 - I_2 = \tfrac{\pi}{4} - \tfrac23.$

Worked Example 4 — the secant reduction (parts and identity, non-zero boundary)

This is the hardest standard reduction and a frequent A* discriminator. For $I_n = \displaystyle\int \sec^n x\,\mathrm{d}x$ , split $\sec^n x = \sec^{n-2}x\cdot\sec^2 x$ and integrate by parts with $u=\sec^{n-2}x,\ \dfrac{\mathrm{d}v}{\mathrm{d}x}=\sec^2 x$ ; then $\dfrac{\mathrm{d}u}{\mathrm{d}x} = (n-2)\sec^{n-2}x\tan x$ (since $\dfrac{\mathrm{d}}{\mathrm{d}x}\sec^{n-2}x = (n-2)\sec^{n-3}x\cdot\sec x\tan x$ ) and $v=\tan x$ :

$I_n = \sec^{n-2}x\,\tan x - (n-2)\int \sec^{n-2}x\,\tan^2 x\,\mathrm{d}x.$

(M1 parts with this split.) Now use $\tan^2 x = \sec^2 x - 1$ :

$I_n = \sec^{n-2}x\,\tan x - (n-2)\int \sec^{n}x\,\mathrm{d}x + (n-2)\int\sec^{n-2}x\,\mathrm{d}x = \sec^{n-2}x\,\tan x - (n-2)I_n + (n-2)I_{n-2}.$

(M1 identity, recovering $I_n$ and $I_{n-2}$ .) Collecting the $I_n$ terms, $(n-1)I_n = \sec^{n-2}x\,\tan x + (n-2)I_{n-2}$ , so

$\boxed{\,I_n = \frac{\sec^{n-2}x\,\tan x}{n-1} + \frac{n-2}{n-1}\,I_{n-2}\,} \qquad (n\ge2).$

(A1 for the recurrence.) For the definite version on $[0,\tfrac{\pi}{4}]$ the boundary term is not zero: with $I_1 = \int_0^{\pi/4}\sec x\,\mathrm{d}x = [\ln|\sec x+\tan x|]_0^{\pi/4} = \ln(\sqrt2+1)$ , the recurrence gives $I_3 = \dfrac{\sec\tfrac{\pi}{4}\tan\tfrac{\pi}{4}}{2} + \dfrac12 I_1 = \dfrac{\sqrt2}{2} + \dfrac12\ln(\sqrt2+1)$ . This example is instructive precisely because it combines all three techniques — parts, a Pythagorean identity, and a non-vanishing boundary term — in one derivation.

4 Specimen-style exam question

(Specimen-style — not a real past paper.)

Let $\displaystyle I_n = \int_0^{1} x^n\,(1-x)^{1/2}\,dx$ . (a) Using integration by parts, show that $\displaystyle I_n = \frac{2n}{2n+3}\,I_{n-1}$ for $n \ge 1$ . (b) Find $I_0$ , and hence evaluate $I_2$ .

(a) Take $u = x^n,\ \dfrac{\mathrm{d}v}{\mathrm{d}x} = (1-x)^{1/2}$ , so $\dfrac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}$ and $v = -\tfrac23(1-x)^{3/2}$ :

$I_n = \Big[ -\tfrac23 x^n (1-x)^{3/2} \Big]_0^{1} + \tfrac{2n}{3}\int_0^{1} x^{n-1}(1-x)^{3/2}\,dx.$

The boundary term is $0$ (at $x=1$ the factor $(1-x)^{3/2}=0$ ; at $x=0$ the factor $x^n = 0$ for $n\ge1$ ). Writing $(1-x)^{3/2} = (1-x)(1-x)^{1/2}$ and expanding,

$I_n = \tfrac{2n}{3}\int_0^{1}\!\big[x^{n-1}(1-x)^{1/2} - x^{n}(1-x)^{1/2}\big]dx = \tfrac{2n}{3}\big(I_{n-1} - I_n\big).$

So $3 I_n = 2n I_{n-1} - 2n I_n$ , giving $(3+2n)I_n = 2n I_{n-1}$ , i.e. $I_n = \dfrac{2n}{2n+3}I_{n-1}$ , as required. $\;\square$

(b) $I_0 = \displaystyle\int_0^{1}(1-x)^{1/2}\,dx = \Big[-\tfrac23(1-x)^{3/2}\Big]_0^{1} = \tfrac23$ . Then $I_1 = \tfrac{2}{5}I_0 = \tfrac{4}{15}$ and $I_2 = \tfrac{4}{7}I_1 = \tfrac{16}{105}$ . Notice how the descent here steps down by one (because the parts peeled off a single power of $x$ ), so only the single base case $I_0$ is needed — contrast the sine recurrence, which steps by two and so requires both $I_0$ and $I_1$ . Reading off the step size from the recurrence tells you exactly how many base cases to compute before you begin the descent.

5 Wallis's formulae and a reduction for $x^n\cos x$

Iterating the sine recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ all the way down gives Wallis's formulae:

$\int_0^{\pi/2}\sin^{2m}x\,dx = \frac{(2m-1)!!}{(2m)!!}\cdot\frac{\pi}{2}, \qquad \int_0^{\pi/2}\sin^{2m+1}x\,dx = \frac{(2m)!!}{(2m+1)!!},$

where the double factorial $k!! = k(k-2)(k-4)\cdots$ runs down in steps of two to $1$ or $2$ . For example $6!! = 6\cdot4\cdot2 = 48$ and $5!! = 5\cdot3\cdot1 = 15$ , so $\int_0^{\pi/2}\sin^{6}x\,dx = \tfrac{5!!}{6!!}\cdot\tfrac{\pi}{2} = \tfrac{15}{48}\cdot\tfrac{\pi}{2} = \tfrac{5\pi}{32}$ — matching Worked Example 2.

These formulae are just the recurrence iterated. For an even index, say $n=2m$ , each step pulls out a factor and lands eventually on $I_0 = \tfrac{\pi}{2}$ :

$I_{2m} = \frac{2m-1}{2m}\cdot\frac{2m-3}{2m-2}\cdots\frac{3}{4}\cdot\frac{1}{2}\cdot I_0,$

and the chain of odd numerators over even denominators is exactly $\dfrac{(2m-1)!!}{(2m)!!}$ . For an odd index the descent ends on $I_1 = 1$ instead, so no $\pi$ survives — which is why $I_3 = \tfrac23,\ I_5 = \tfrac{8}{15}$ are purely rational. Recognising whether your target index is even or odd tells you immediately whether to expect a $\pi$ in the answer, a quick check that catches many slips.

A reduction can also reduce a power by two while leaving explicit boundary terms. For the indefinite integral $I_n = \displaystyle\int x^n\cos x\,dx$ , parts with $u=x^n,\ \tfrac{\mathrm{d}v}{\mathrm{d}x}=\cos x$ gives $I_n = x^n\sin x - n\int x^{n-1}\sin x\,dx$ ; a second application to $\int x^{n-1}\sin x\,dx$ yields

$I_n = x^n\sin x + n\,x^{n-1}\cos x - n(n-1)\,I_{n-2}.$

Reduction Formulae

Reduction Formulae

1 Spec mapping (AQA 7367)

2 Core theory — what a reduction formula is, and why it works

3 Worked examples with M1/A1 mark scheme

Worked Example 1 — In=∫01xnex dx\displaystyle I_n = \int_0^{1} x^n e^{x}\,dxIn​=∫01​xnexdx

Worked Example 2 — the sine reduction In=∫0π/2sin⁡nx dx\displaystyle I_n = \int_0^{\pi/2} \sin^n x\,dxIn​=∫0π/2​sinnxdx

Worked Example 3 — the tangent reduction In=∫0π/4tan⁡nx dx\displaystyle I_n = \int_0^{\pi/4} \tan^n x\,dxIn​=∫0π/4​tannxdx

Worked Example 4 — the secant reduction (parts and identity, non-zero boundary)

4 Specimen-style exam question

5 Wallis's formulae and a reduction for xncos⁡xx^n\cos xxncosx

More in Mathematics

Worked Example 1 — $\displaystyle I_n = \int_0^{1} x^n e^{x}\,dx$

Worked Example 2 — the sine reduction $\displaystyle I_n = \int_0^{\pi/2} \sin^n x\,dx$

Worked Example 3 — the tangent reduction $\displaystyle I_n = \int_0^{\pi/4} \tan^n x\,dx$

5 Wallis's formulae and a reduction for $x^n\cos x$