What is wrong base case?

sin^n and cos^n reductions step by two, so an even n ends at I_0 and an odd n at I_1; using the wrong base case gives the wrong constant.

What are boundary term errors?

Always evaluate the [uv] term; for 0 to π2 it usually vanishes, but check rather than assume.

What is sign slip from the Pythagorean substitution?

sin^2 x = 1 - cos^2 x introduces -I_n; mishandling the sign breaks the rearrangement.

State the reduction formula for I_n = _0^π/2sin^n x\,dx. [1 mark]

What is the base case I_0 for _0^π/2sin^n x\,dx? [1 mark]

Using I_n = e - n I_n-1 with I_0 = e - 1, find I_1. [1 mark]

SEAB Original-style practice: Let I_n = _0^1 x^n e^x\,dx. Show that I_n = e - n I_n-1 for n ≥ 1, and hence evaluate I_2.

Apply integration by parts to I_n = _0^1 x^n e^x\,dx with u = x^n (so du = nx^n-1dx) and dv = e^x dx (so v = e^x): $I_n = [x^n e^x]_0^1 - _0^1 n x^n-1e^x\,dx = (1·e - 0) - n_0^1 x^n-1e^x\,dx = e - n I_n-1.$ Base case: I_0 = _0^1 e^x\,dx = [e^x]_0^1 = e - 1. Then I_1 = e - 1· I_0 = e - (e - 1) = 1. And I_2 = e - 2 I_1 = e - 2. So I_2 = e - 2. Markers reward the by-parts choice, the boundary term and reduced integral giving e - n I_n-1, the base case I_0 = e - 1, and the value I_2 = e - 2.

SEAB Original-style practice: Given I_n = _0^π/2 sin^n x\,dx, the reduction formula I_n = n-1nI_n-2 holds for n ≥ 2. Use it to evaluate _0^π/2sin^4 x\,dx.

Apply the reduction formula twice down to a base case. The base case is I_0 = _0^π/2 1\,dx = π2. I_4 = 4 - 14 I_2 = 34I_2. And I_2 = 2 - 12I_0 = 12·π2 = π4. Therefore I_4 = 34·π4 = 3π16. So _0^π/2sin^4 x\,dx = 3π16. Markers reward stepping down with the reduction formula, the base case I_0 = π2, computing I_2 = π4, and the final value 3π16.

SingaporeFurther MathsSyllabus dot point

How do we derive and use a reduction formula to integrate a family of integrals indexed by an integer?

Derive reduction formulae using integration by parts and apply them to evaluate families of integrals

A focused answer to the H2 Further Mathematics outcome on reduction formulae. Deriving a recurrence for an integral by integration by parts, applying it repeatedly down to a base case, and standard reduction formulae for powers of sine and cosine.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to derive a reduction formula, a recurrence that expresses an integral $I_n$ in terms of an integral of lower index, using integration by parts, and then to apply it repeatedly down to a base case to evaluate a specific integral. Reduction formulae are the systematic way to integrate powers of functions that resist a single direct method.

The answer

What a reduction formula is

A reduction formula expresses an integral indexed by an integer $n$ , say $I_n = \displaystyle\int x^n\mathrm{f}(x)\,dx$ , in terms of $I_{n-1}$ or $I_{n-2}$ . Repeated application steps the index down until it reaches a base case ( $I_0$ or $I_1$ ) that can be integrated directly.

Deriving one by parts

The standard derivation splits off one factor and integrates by parts. For $I_n = \int x^n\mathrm{e}^x\,dx$ , take $u = x^n$ and $dv = \mathrm{e}^x\,dx$ ; integration by parts lowers the power of $x$ by one, producing $I_{n-1}$ . For powers of sine, $\sin^n x = \sin^{n-1}x\cdot\sin x$ , integrate by parts with $dv = \sin x\,dx$ , then use $\cos^2 x = 1 - \sin^2 x$ to express the result back in terms of $I_n$ and $I_{n-2}$ , and rearrange.

The rearrangement step

For trigonometric powers the by-parts result contains $I_n$ again on the right. Collect the $I_n$ terms on the left and divide. For $\int_0^{\pi/2}\sin^n x\,dx$ this gives the clean

I_n = \frac{n-1}{n}\,I_{n-2}.

Applying the formula

Identify the base case, then step down. For a definite integral the boundary terms often vanish (for example $[\sin^{n-1}x\cos x]_0^{\pi/2} = 0$ ), which is why the symmetric limits $0$ to $\tfrac{\pi}{2}$ give such a tidy recurrence.

Worked example

Let $I_n = \displaystyle\int_0^{\pi/2}\cos^n x\,dx$ . Derive the reduction formula $I_n = \dfrac{n-1}{n}I_{n-2}$ for $n \geq 2$ , then find $I_5$ .

Step 1: Split and integrate by parts

Write $\cos^n x = \cos^{n-1}x\cdot\cos x$ . Take $u = \cos^{n-1}x$ and $dv = \cos x\,dx$ , so $du = -(n-1)\cos^{n-2}x\sin x\,dx$ and $v = \sin x$ :

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

Step 2: Evaluate the boundary term

At $x = \tfrac{\pi}{2}$ , $\cos = 0$ ; at $x = 0$ , $\sin = 0$ . So the boundary term is $0$ .

Step 3: Use the Pythagorean identity

Replace $\sin^2 x = 1 - \cos^2 x$ :

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

Step 4: Collect and rearrange

I_n = (n-1)I_{n-2} - (n-1)I_n \;\Rightarrow\; I_n + (n-1)I_n = (n-1)I_{n-2} \;\Rightarrow\; n I_n = (n-1)I_{n-2},

so $I_n = \dfrac{n-1}{n}I_{n-2}$ .

Step 5: Apply to find I5

Base case $I_1 = \int_0^{\pi/2}\cos x\,dx = [\sin x]_0^{\pi/2} = 1$ . Then $I_3 = \tfrac{2}{3}I_1 = \tfrac{2}{3}$ , and $I_5 = \tfrac{4}{5}I_3 = \tfrac{4}{5}\cdot\tfrac{2}{3} = \tfrac{8}{15}$ .

Common traps

Not collecting the repeated $I_n$: Forgetting to gather $I_n$ terms leaves an unusable circular identity; the rearrangement is essential.
Wrong base case: $\sin^n$ and $\cos^n$ reductions step by two, so an even $n$ ends at $I_0$ and an odd $n$ at $I_1$ ; using the wrong base case gives the wrong constant.
Boundary term errors: Always evaluate the $[uv]$ term; for $0$ to $\tfrac{\pi}{2}$ it usually vanishes, but check rather than assume.
Stopping at the wrong index: Apply the formula until the index reaches the base case; stopping early leaves an unevaluated integral.
Sign slip from the Pythagorean substitution: $\sin^2 x = 1 - \cos^2 x$ introduces $-I_n$ ; mishandling the sign breaks the rearrangement.

Examples in context

Example 1. Wallis's product. Repeatedly applying the $\sin^n$ reduction formula and comparing even and odd cases yields Wallis's famous product for $\pi$ , a classic demonstration that reduction formulae are not just a computational trick but a route to deep results.

Example 2. Moments of a distribution. Integrals such as $\int x^n\mathrm{e}^{-x}\,dx$ that define the moments of an exponential-type distribution are evaluated by a reduction formula, linking the technique directly to expectation calculations in statistics.

Try this

Q1. State the reduction formula for $I_n = \int_0^{\pi/2}\sin^n x\,dx$ . [1 mark]

Cue. $I_n = \dfrac{n-1}{n}I_{n-2}$ for $n \geq 2$ .

Q2. What is the base case $I_0$ for $\int_0^{\pi/2}\sin^n x\,dx$ ? [1 mark]

Cue. $I_0 = \int_0^{\pi/2} 1\,dx = \dfrac{\pi}{2}$ .

Q3. Using $I_n = \mathrm{e} - n I_{n-1}$ with $I_0 = \mathrm{e} - 1$ , find $I_1$ . [1 mark]

Cue. $I_1 = \mathrm{e} - 1\cdot(\mathrm{e} - 1) = 1$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original7 marksLet

I_n = \displaystyle\int_0^{1} x^n \mathrm{e}^{x}\,dx

. Show that

I_n = \mathrm{e} - n I_{n-1}

for

n \geq 1

, and hence evaluate

I_2

Show worked answer →

Apply integration by parts to $I_n = \int_0^1 x^n \mathrm{e}^x\,dx$ with $u = x^n$ (so $du = nx^{n-1}dx$ ) and $dv = \mathrm{e}^x dx$ (so $v = \mathrm{e}^x$ ):

I_n = \left[x^n \mathrm{e}^x\right]_0^1 - \int_0^1 n x^{n-1}\mathrm{e}^x\,dx = (1\cdot\mathrm{e} - 0) - n\int_0^1 x^{n-1}\mathrm{e}^x\,dx = \mathrm{e} - n I_{n-1}.

Base case:

I_0 = \int_0^1 \mathrm{e}^x\,dx = [\mathrm{e}^x]_0^1 = \mathrm{e} - 1

Then $I_1 = \mathrm{e} - 1\cdot I_0 = \mathrm{e} - (\mathrm{e} - 1) = 1$ . And $I_2 = \mathrm{e} - 2 I_1 = \mathrm{e} - 2$ .

So $I_2 = \mathrm{e} - 2$ .

Markers reward the by-parts choice, the boundary term and reduced integral giving $\mathrm{e} - n I_{n-1}$ , the base case $I_0 = \mathrm{e} - 1$ , and the value $I_2 = \mathrm{e} - 2$ .

Original6 marksGiven

I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,dx

, the reduction formula

I_n = \dfrac{n-1}{n}I_{n-2}

holds for

n \geq 2

. Use it to evaluate

\displaystyle\int_0^{\pi/2}\sin^4 x\,dx

Show worked answer →

Apply the reduction formula twice down to a base case. The base case is $I_0 = \int_0^{\pi/2} 1\,dx = \dfrac{\pi}{2}$ .

$I_4 = \dfrac{4 - 1}{4} I_2 = \dfrac{3}{4}I_2$ . And $I_2 = \dfrac{2 - 1}{2}I_0 = \dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi}{4}$ .

Therefore $I_4 = \dfrac{3}{4}\cdot\dfrac{\pi}{4} = \dfrac{3\pi}{16}$ .

So $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx = \dfrac{3\pi}{16}$ .

Markers reward stepping down with the reduction formula, the base case $I_0 = \tfrac{\pi}{2}$ , computing $I_2 = \tfrac{\pi}{4}$ , and the final value $\tfrac{3\pi}{16}$ .