Home / A-Level Further Maths 9231 / Further integration / Reduction formulae
Cambridge International AS & A Level · Topic 2.4
A-Level Further Maths 9231 / Topic 2.4 / AS & A Level

One integral, a whole ladder of them.

A reduction formula links \(I_n\) to \(I_{n-1}\) or \(I_{n-2}\), so one integration by parts handles every power at once. Derive formulae by splitting off a factor, by rewriting \(\cos^{2}x\) as \(1 - \sin^{2}x\), or by differentiating a product you are given, then climb down the ladder to \(I_0\) or \(I_1\) for an exact value.

Topic 2.4All further integration lessons →
The key idea

Write \(I_n\) for a definite integral that depends on a whole number \(n\). A reduction formula expresses \(I_n\) in terms of \(I_{n-1}\) or \(I_{n-2}\). Derive it by integration by parts (split off one factor so the power drops), by an identity such as \(\cos^{2}x = 1 - \sin^{2}x\) that brings \(I_n\) back on the right, or by integrating a given derivative between the limits. Evaluate every boundary term carefully, then apply the formula repeatedly down to \(I_0\) or \(I_1\), which you integrate directly. For example \(\displaystyle\int_0^{\pi/2}\sin^{n}x\,dx = \frac{n-1}{n}I_{n-2}\).

x y π/2 0 1 area under y = sinⁿ x n = 1: 1 n = 2: 0.785 n = 4: 0.589 n = 8: 0.430 each step down multiplies by (n − 1)/n
Fig. 1: the graphs of y = sin^n x on 0 to pi/2 for n = 1, 2, 4 and 8. Every curve passes through (pi/2, 1), but higher powers hug the axis for longer, so the area I_n shrinks. The reduction formula I_n = ((n - 1)/n) I_(n - 2) gives each area from the one two steps below it
Section 01

Split off a factor and use parts.

Write \(I_n\) for the integral with power \(n\). Choose \(u\) in integration by parts so that differentiating it lowers the power: \(u = x^{n}\) in \(x^{n}e^{x}\), \(u = (\ln x)^{n}\) in \((\ln x)^{n}\), \(u = x^{n}\) in \(x^{n}\sqrt{1 - x}\). The new integral is then \(I_{n-1}\) (or something you can rewrite in terms of \(I_{n-1}\) and \(I_n\)), and the boundary term is just numbers.

Three habits make a derivation earn full marks:

HabitWhy
Evaluate the boundary term \(\big[uv\big]_a^b\) at both limitsit is usually where the constant (\(e\), \(2^{n}\), \(\sqrt{2}\)) comes from, and often one end gives 0
State the values of \(n\) for which it holds\(x^{n-1}\) or \(\sin^{n-2}x\) only makes sense, and a boundary term only vanishes, for large enough \(n\)
If \(I_n\) reappears on the right, collect itmove all the \(I_n\) terms to the left and divide

Then find \(I_0\) (or \(I_1\)) by direct integration and step up: \(I_0 \to I_1 \to I_2 \to \dots\). Keep each value exact.

Worked example A formula and a value
Given \(I_n = \displaystyle\int_0^1 x^{n}e^{x}\,dx\), show that \(I_n = e - nI_{n-1}\) for \(n \ge 1\). Hence find the exact value of \(I_3\).
  1. Take \(u = x^{n}\), \(\dfrac{dv}{dx} = e^{x}\), so \(\dfrac{du}{dx} = nx^{n-1}\) and \(v = e^{x}\): \[\begin{aligned} I_n &= \Big[x^{n}e^{x}\Big]_0^1 - \int_0^1 nx^{n-1}e^{x}\,dx \\ &= (e - 0) - nI_{n-1}. \end{aligned}\]
  2. At \(x = 0\) the term is \(0^{n}e^{0} = 0\) because \(n \ge 1\). So \(I_n = e - nI_{n-1}\).
  3. \(I_0 = \displaystyle\int_0^1 e^{x}\,dx = e - 1\).
  4. \(I_1 = e - (e - 1) = 1\), \(I_2 = e - 2 \times 1 = e - 2\).
  5. \(I_3 = e - 3(e - 2) = 6 - 2e\).
Answer: \(I_n = e - nI_{n-1}\) and \(I_3 = 6 - 2e\) (about 0.563)
Worked example When I_n comes back
Given \(I_n = \displaystyle\int_0^1 x^{n}\sqrt{1 - x}\,dx\), show that \((2n + 3)I_n = 2nI_{n-1}\) for \(n \ge 1\), and find \(I_2\).
  1. Take \(u = x^{n}\), \(\dfrac{dv}{dx} = (1 - x)^{1/2}\), so \(v = -\tfrac{2}{3}(1 - x)^{3/2}\): \[\begin{aligned} I_n &= \Big[-\tfrac{2}{3}x^{n}(1 - x)^{3/2}\Big]_0^1 \\ &\quad + \tfrac{2n}{3}\int_0^1 x^{n-1}(1 - x)^{3/2}\,dx. \end{aligned}\]
  2. The bracket is 0 at both ends (\(1 - x = 0\) at the top, \(x^{n} = 0\) at the bottom).
  3. Write \((1 - x)^{3/2} = (1 - x)(1 - x)^{1/2}\): \[\begin{aligned} \int_0^1 x^{n-1}(1 - x)^{3/2}\,dx &= \int_0^1 (x^{n-1} - x^{n})\sqrt{1 - x}\,dx \\ &= I_{n-1} - I_n. \end{aligned}\]
  4. So \(I_n = \tfrac{2n}{3}(I_{n-1} - I_n)\), giving \(3I_n + 2nI_n = 2nI_{n-1}\), that is \((2n + 3)I_n = 2nI_{n-1}\).
  5. \(I_0 = \displaystyle\int_0^1 (1 - x)^{1/2}\,dx = \tfrac{2}{3}\). Then \(I_1 = \tfrac{2}{5} \times \tfrac{2}{3} = \tfrac{4}{15}\) and \(I_2 = \tfrac{4}{7} \times \tfrac{4}{15} = \tfrac{16}{105}\).
Answer: \((2n + 3)I_n = 2nI_{n-1}\), and \(I_2 = \tfrac{16}{105}\)
Your turn
Given \(I_n = \displaystyle\int_1^e (\ln x)^{n}\,dx\), show that \(I_n = e - nI_{n-1}\) for \(n \ge 1\) and find \(I_3\).
Examiner trap

Do not drop the boundary term or evaluate it at one end only. In \(\big[x^{n}e^{x}\big]_0^1\) the top gives \(e\) and the bottom gives 0 only because \(n \ge 1\); writing \(I_n = 1 - nI_{n-1}\) (from \(e^{0}\)) or forgetting the minus sign in front of the new integral ruins every value further up the ladder. And \(I_0\) is \(e - 1\), not \(e\).

Section 02

Powers of sine and cosine.

For \(I_n = \displaystyle\int_0^{\pi/2} \sin^{n}x\,dx\), split off one sine: \(\sin^{n}x = \sin^{n-1}x \times \sin x\). Integrating by parts leaves \(\sin^{n-2}x\cos^{2}x\), and \(\cos^{2}x = 1 - \sin^{2}x\) turns that into \(I_{n-2} - I_n\). Collect the \(I_n\) terms and divide by \(n\).

The result, for \(n \ge 2\), is the same for sine and cosine on this interval:

\[\int_0^{\pi/2}\sin^{n}x\,dx = \frac{n-1}{n}\,I_{n-2}, \qquad \int_0^{\pi/2}\cos^{n}x\,dx = \frac{n-1}{n}\,J_{n-2}.\]

Where the ladder endsValueExample
\(n\) odd: down to \(I_1\)\(\displaystyle\int_0^{\pi/2}\sin x\,dx = 1\)\(I_5 = \tfrac{4}{5} \times \tfrac{2}{3} \times 1 = \tfrac{8}{15}\)
\(n\) even: down to \(I_0\)\(\displaystyle\int_0^{\pi/2}1\,dx = \tfrac{\pi}{2}\)\(I_4 = \tfrac{3}{4} \times \tfrac{1}{2} \times \tfrac{\pi}{2} = \tfrac{3\pi}{16}\)

So an odd power gives a rational answer and an even power gives a rational multiple of \(\pi\). That is a quick check on any value you find.

Worked example Deriving the sine formula
Given \(I_n = \displaystyle\int_0^{\pi/2}\sin^{n}x\,dx\), show that \(nI_n = (n - 1)I_{n-2}\) for \(n \ge 2\).
  1. Take \(u = \sin^{n-1}x\), \(\dfrac{dv}{dx} = \sin x\), so \(\dfrac{du}{dx} = (n - 1)\sin^{n-2}x\cos x\) and \(v = -\cos x\): \[\begin{aligned} I_n &= \Big[-\sin^{n-1}x\cos x\Big]_0^{\pi/2} \\ &\quad + (n - 1)\int_0^{\pi/2}\sin^{n-2}x\cos^{2}x\,dx. \end{aligned}\]
  2. Boundary term: \(\cos\tfrac{\pi}{2} = 0\) at the top and \(\sin 0 = 0\) at the bottom (since \(n - 1 \ge 1\)), so it is 0.
  3. Use \(\cos^{2}x = 1 - \sin^{2}x\): \[\begin{aligned} I_n &= (n - 1)\int_0^{\pi/2}\big(\sin^{n-2}x - \sin^{n}x\big)\,dx \\ &= (n - 1)I_{n-2} - (n - 1)I_n. \end{aligned}\]
  4. Collect: \(I_n + (n - 1)I_n = (n - 1)I_{n-2}\), so \(nI_n = (n - 1)I_{n-2}\).
Answer: \(nI_n = (n - 1)I_{n-2}\), that is \(I_n = \dfrac{n-1}{n}I_{n-2}\) for \(n \ge 2\)
Worked example Climbing down to the base
Using \(I_n = \dfrac{n-1}{n}I_{n-2}\), find \(\displaystyle\int_0^{\pi/2}\sin^{6}x\,dx\) and \(\displaystyle\int_0^{\pi/2}\sin^{5}x\,dx\) exactly.
  1. Even power, so go down to \(I_0 = \tfrac{\pi}{2}\): \[I_6 = \tfrac{5}{6}I_4 = \tfrac{5}{6}\cdot\tfrac{3}{4}I_2 = \tfrac{5}{6}\cdot\tfrac{3}{4}\cdot\tfrac{1}{2}I_0.\]
  2. \(I_6 = \tfrac{15}{48} \times \tfrac{\pi}{2} = \tfrac{5\pi}{32}\).
  3. Odd power, so go down to \(I_1 = \big[-\cos x\big]_0^{\pi/2} = 1\): \[I_5 = \tfrac{4}{5}I_3 = \tfrac{4}{5}\cdot\tfrac{2}{3}I_1 = \tfrac{8}{15}.\]
  4. Check: \(I_6 \approx 0.491 \lt I_5 \approx 0.533\), as it should be, since \(\sin^{6}x \le \sin^{5}x\) on the interval.
Answer: \(I_6 = \dfrac{5\pi}{32}\) and \(I_5 = \dfrac{8}{15}\)
Your turn
Find \(\displaystyle\int_0^{\pi/2}\cos^{4}x\,dx\) exactly, using the cosine version of the formula.
Examiner trap

The fraction is \(\dfrac{n-1}{n}\), not \(\dfrac{n}{n-1}\): every factor is less than 1, so the integrals get smaller as \(n\) grows (see Fig. 1). Stop at the right base: \(I_0 = \tfrac{\pi}{2}\) for even \(n\), \(I_1 = 1\) for odd \(n\). Using \(I_0 = 1\) loses the \(\pi\), and going one step too few gives \(I_{n-2}\) instead of \(I_n\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

For \(I_n = \displaystyle\int_0^1 x^{n}e^{x}\,dx\) and \(n \ge 1\), which reduction formula is correct?

Quick check+10 XP

Using \(I_n = \dfrac{n-1}{n}I_{n-2}\) for \(I_n = \displaystyle\int_0^{\pi/2}\sin^{n}x\,dx\), find \(I_5\).

Quick check+10 XP

In deriving the formula for \(\displaystyle\int_0^{\pi/2}\sin^{n}x\,dx\) with \(n \ge 2\), what is the value of \(\Big[-\sin^{n-1}x\cos x\Big]_0^{\pi/2}\)?

Quick check+10 XP

Given \(I_n = \displaystyle\int_0^{\pi/4}\tan^{n}x\,dx\), \(I_n + I_{n-2} = \dfrac{1}{n - 1}\) and \(I_0 = \tfrac{\pi}{4}\), find \(I_2\).

Section 03

Using a hint: differentiate a product.

In harder cases the question gives you a product to differentiate, such as \(x(1 + x^{2})^{n}\) or \(\tan^{n-1}x\). Differentiate it, rewrite the derivative in terms of the integrands of \(I_n\) and \(I_{n-1}\) (or \(I_{n-2}\)), then integrate both sides between the limits. The left side becomes the product evaluated at the limits, so no integration by parts is needed.

The pattern, when \(\dfrac{d}{dx}\big(P(x)\big) = A f_n(x) + B f_{n-1}(x)\):

\[\Big[P(x)\Big]_a^b = A\,I_n + B\,I_{n-1}.\]

The algebra in the middle is usually one identity: \(x^{2} = (1 + x^{2}) - 1\), \(\tan^{2}x = \sec^{2}x - 1\), or \(\cos^{2}x = 1 - \sin^{2}x\). Use it so that every term is a power of the SAME expression as the integrand.

Worked example A hint with a power of 2
Given \(I_n = \displaystyle\int_0^1 (1 + x^{2})^{n}\,dx\), by considering \(\dfrac{d}{dx}\big(x(1 + x^{2})^{n}\big)\) show that \((2n + 1)I_n = 2^{n} + 2nI_{n-1}\) for \(n \ge 1\). Hence find \(I_2\).
  1. Product rule: \[\begin{aligned} \frac{d}{dx}\big(x(1 + x^{2})^{n}\big) &= (1 + x^{2})^{n} + 2nx^{2}(1 + x^{2})^{n-1}. \end{aligned}\]
  2. Write \(x^{2} = (1 + x^{2}) - 1\): \[\begin{aligned} 2nx^{2}(1 + x^{2})^{n-1} &= 2n(1 + x^{2})^{n} \\ &\quad - 2n(1 + x^{2})^{n-1}. \end{aligned}\]
  3. So the derivative is \((2n + 1)(1 + x^{2})^{n} - 2n(1 + x^{2})^{n-1}\).
  4. Integrate from 0 to 1: \(\big[x(1 + x^{2})^{n}\big]_0^1 = 2^{n} - 0\), so \(2^{n} = (2n + 1)I_n - 2nI_{n-1}\).
  5. \(I_0 = 1\). \(n = 1\): \(3I_1 = 2 + 2 = 4\), \(I_1 = \tfrac{4}{3}\). \(n = 2\): \(5I_2 = 4 + 4 \times \tfrac{4}{3} = \tfrac{28}{3}\), so \(I_2 = \tfrac{28}{15}\).
  6. Check by expanding: \(\displaystyle\int_0^1 (1 + 2x^{2} + x^{4})\,dx = 1 + \tfrac{2}{3} + \tfrac{1}{5} = \tfrac{28}{15}\).
Answer: \((2n + 1)I_n = 2^{n} + 2nI_{n-1}\), and \(I_2 = \tfrac{28}{15}\)
Worked example Powers of tan
Given \(I_n = \displaystyle\int_0^{\pi/4}\tan^{n}x\,dx\), by considering \(\dfrac{d}{dx}\big(\tan^{n-1}x\big)\) show that \(I_n + I_{n-2} = \dfrac{1}{n - 1}\) for \(n \ge 2\). Hence find \(I_4\).
  1. Chain rule: \(\dfrac{d}{dx}\big(\tan^{n-1}x\big) = (n - 1)\tan^{n-2}x\sec^{2}x\).
  2. Write \(\sec^{2}x = 1 + \tan^{2}x\): \[\frac{d}{dx}\big(\tan^{n-1}x\big) = (n - 1)\big(\tan^{n}x + \tan^{n-2}x\big).\]
  3. Integrate from 0 to \(\tfrac{\pi}{4}\): \(\big[\tan^{n-1}x\big]_0^{\pi/4} = 1 - 0 = 1\) (as \(n - 1 \ge 1\)), so \(1 = (n - 1)(I_n + I_{n-2})\).
  4. \(I_0 = \tfrac{\pi}{4}\). \(n = 2\): \(I_2 = 1 - \tfrac{\pi}{4}\). \(n = 4\): \(I_4 = \tfrac{1}{3} - I_2 = \tfrac{\pi}{4} - \tfrac{2}{3}\).
Answer: \(I_n + I_{n-2} = \dfrac{1}{n - 1}\), and \(I_4 = \dfrac{\pi}{4} - \dfrac{2}{3}\) (about 0.119)
Your turn
With \(I_n = \displaystyle\int_0^{\pi/4}\tan^{n}x\,dx\) and \(I_n + I_{n-2} = \dfrac{1}{n - 1}\), find \(I_3\) exactly.
Examiner trap

When you integrate the derivative, the left side is the product evaluated at the limits, so it carries the constant: \(\big[x(1 + x^{2})^{n}\big]_0^1 = 2^{n}\), not 0 and not 1. Leaving it out gives a 'formula' with no constant that produces wrong values. Also rewrite with the right identity: \(\sec^{2}x = 1 + \tan^{2}x\), not \(1 - \tan^{2}x\).

Stage 2 · Exam

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Given \(I_n = \displaystyle\int_0^1 x^{n}e^{x}\,dx\) with \(I_n = e - nI_{n-1}\), find the exact value of \(\displaystyle\int_0^1 x^{3}e^{x}\,dx\).

Exam style+20 XP

Find the exact value of \(\displaystyle\int_0^{\pi/2}\cos^{6}x\,dx\).

Exam style+20 XP

Given \(I_n = \displaystyle\int_0^1 x^{n}\sqrt{1 - x}\,dx\) satisfies \((2n + 3)I_n = 2nI_{n-1}\), find \(I_2\).

Exam style+20 XP

Given \(I_n = \displaystyle\int_0^1 (1 + x^{2})^{n}\,dx\) and \(\dfrac{d}{dx}\big(x(1 + x^{2})^{n}\big) = (2n + 1)(1 + x^{2})^{n} - 2n(1 + x^{2})^{n-1}\), find \(I_2\).

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Six original Cambridge-style questions
Model derivations by parts for \(x^{n}e^{2x}\), \(x^{n}\sin x\) and \(x^{n}\sqrt{1 + x}\), the cosine formula with a product of powers, and two hinted formulae for \((1 - x^{2})^{n}\) and \(\sec^{n}x\), each evaluated exactly.