AS & A Level · Practice questions · Reduction formulae

Derive it, then climb down.

Six original Cambridge-style questions on reduction formulae: integration by parts once and twice, an integral that reappears on the right, the \(\cos^{n}x\) formula, and hints given as a product to differentiate, with full model solutions and exact values.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Split, collect, then step down.

Split off one factor so that differentiating \(u\) lowers the power: \(x^{n}\), \((\ln x)^{n}\), \(\sin^{n-1}x\).
Evaluate the boundary term at BOTH limits and state the values of \(n\) for which it holds.
Rewrite with an identity (\(\cos^{2}x = 1 - \sin^{2}x\), \(x^{2} = (1 + x^{2}) - 1\)) so only \(I_n\) and \(I_{n-1}\) or \(I_{n-2}\) appear, then collect the \(I_n\) terms.
Hint given? Differentiate the product, rewrite, and integrate both sides between the limits.
Step from \(I_0\) or \(I_1\), found directly, up to the \(I_n\) you need, keeping every value exact.

Recall

[4 marks]

Given \(I_n = \displaystyle\int_0^1 x^{n}e^{2x}\,dx\), show that \(I_n = \tfrac{1}{2}e^{2} - \tfrac{n}{2}I_{n-1}\) for \(n \ge 1\). Hence find the exact value of \(I_3\).

Take \(u = x^{n}\), \(\dfrac{dv}{dx} = e^{2x}\), so \(v = \tfrac{1}{2}e^{2x}\): \[\begin{aligned} I_n &= \Big[\tfrac{1}{2}x^{n}e^{2x}\Big]_0^1 - \tfrac{n}{2}\int_0^1 x^{n-1}e^{2x}\,dx \\ &= \tfrac{1}{2}e^{2} - \tfrac{n}{2}I_{n-1}, \end{aligned}\] the lower limit giving 0 since \(n \ge 1\). ✓
\(I_0 = \big[\tfrac{1}{2}e^{2x}\big]_0^1 = \tfrac{1}{2}(e^{2} - 1)\). ✓
\(I_1 = \tfrac{1}{2}e^{2} - \tfrac{1}{4}(e^{2} - 1) = \tfrac{1}{4}(e^{2} + 1)\). ✓
\(I_2 = \tfrac{1}{2}e^{2} - \tfrac{2}{2} \times \tfrac{1}{4}(e^{2} + 1) = \tfrac{1}{4}(e^{2} - 1)\). ✓
\(I_3 = \tfrac{1}{2}e^{2} - \tfrac{3}{2} \times \tfrac{1}{4}(e^{2} - 1) = \tfrac{1}{8}(e^{2} + 3)\). ✓

Calculation

[5 marks]

Given \(J_n = \displaystyle\int_0^{\pi/2}\cos^{n}x\,dx\), it can be shown that \(J_n = \dfrac{n-1}{n}J_{n-2}\) for \(n \ge 2\).

(a) Find \(J_7\) exactly.

(b) By writing \(\sin^{2}x = 1 - \cos^{2}x\), find the exact value of \(\displaystyle\int_0^{\pi/2}\sin^{2}x\cos^{4}x\,dx\).

(a) Odd power, so go down to \(J_1 = \big[\sin x\big]_0^{\pi/2} = 1\): \(J_7 = \tfrac{6}{7}\cdot\tfrac{4}{5}\cdot\tfrac{2}{3}J_1 = \tfrac{48}{105} = \tfrac{16}{35}\). ✓
(b) \(\sin^{2}x\cos^{4}x = (1 - \cos^{2}x)\cos^{4}x = \cos^{4}x - \cos^{6}x\), so the integral is \(J_4 - J_6\). ✓
\(J_4 = \tfrac{3}{4}\cdot\tfrac{1}{2}\cdot\tfrac{\pi}{2} = \tfrac{3\pi}{16}\) and \(J_6 = \tfrac{5}{6}J_4 = \tfrac{5\pi}{32}\). ✓
\(J_4 - J_6 = \tfrac{6\pi}{32} - \tfrac{5\pi}{32} = \tfrac{\pi}{32}\). ✓

Calculation

[6 marks]

Given \(I_n = \displaystyle\int_0^{\pi}x^{n}\sin x\,dx\), use integration by parts twice to show that \(I_n = \pi^{n} - n(n - 1)I_{n-2}\) for \(n \ge 2\). Hence find \(I_4\) exactly.

First, \(u = x^{n}\), \(v = -\cos x\): \[\begin{aligned} I_n &= \Big[-x^{n}\cos x\Big]_0^{\pi} + n\int_0^{\pi}x^{n-1}\cos x\,dx \\ &= \pi^{n} + n\int_0^{\pi}x^{n-1}\cos x\,dx, \end{aligned}\] since \(-\pi^{n}\cos\pi = \pi^{n}\) and the lower limit gives 0. ✓
Second, \(u = x^{n-1}\), \(v = \sin x\): \[\begin{aligned} \int_0^{\pi}x^{n-1}\cos x\,dx &= \Big[x^{n-1}\sin x\Big]_0^{\pi} \\ &\quad - (n - 1)I_{n-2} \\ &= -(n - 1)I_{n-2}, \end{aligned}\] as \(\sin\pi = \sin 0 = 0\). ✓
So \(I_n = \pi^{n} - n(n - 1)I_{n-2}\). ✓
\(I_0 = \big[-\cos x\big]_0^{\pi} = 2\). \(I_2 = \pi^{2} - 2 \times 1 \times 2 = \pi^{2} - 4\). ✓
\(I_4 = \pi^{4} - 4 \times 3(\pi^{2} - 4) = \pi^{4} - 12\pi^{2} + 48\) (about 26.97). ✓

Proof

[6 marks]

Given \(I_n = \displaystyle\int_0^1 x^{n}\sqrt{1 + x}\,dx\), show that \((2n + 3)I_n = 4\sqrt{2} - 2nI_{n-1}\) for \(n \ge 1\). Hence find \(I_1\) exactly.

Take \(u = x^{n}\), \(\dfrac{dv}{dx} = (1 + x)^{1/2}\), so \(v = \tfrac{2}{3}(1 + x)^{3/2}\): \[\begin{aligned} I_n &= \Big[\tfrac{2}{3}x^{n}(1 + x)^{3/2}\Big]_0^1 \\ &\quad - \tfrac{2n}{3}\int_0^1 x^{n-1}(1 + x)^{3/2}\,dx. \end{aligned}\] ✓
Boundary term: \(\tfrac{2}{3} \times 2^{3/2} - 0 = \tfrac{4\sqrt{2}}{3}\). ✓
\((1 + x)^{3/2} = (1 + x)\sqrt{1 + x}\), so the integral is \(\int_0^1 (x^{n-1} + x^{n})\sqrt{1 + x}\,dx = I_{n-1} + I_n\). ✓
\(I_n = \tfrac{4\sqrt{2}}{3} - \tfrac{2n}{3}(I_{n-1} + I_n)\). Multiply by 3 and collect: \((2n + 3)I_n = 4\sqrt{2} - 2nI_{n-1}\). ✓
\(I_0 = \big[\tfrac{2}{3}(1 + x)^{3/2}\big]_0^1 = \tfrac{2}{3}(2\sqrt{2} - 1)\). ✓
\(n = 1\): \(5I_1 = 4\sqrt{2} - \tfrac{4}{3}(2\sqrt{2} - 1) = \tfrac{4\sqrt{2} + 4}{3}\), so \(I_1 = \tfrac{4}{15}(\sqrt{2} + 1)\). ✓

Analysis

[6 marks]

Given \(I_n = \displaystyle\int_0^1 (1 - x^{2})^{n}\,dx\).

(a) By considering \(\dfrac{d}{dx}\big(x(1 - x^{2})^{n}\big)\), show that \((2n + 1)I_n = 2nI_{n-1}\) for \(n \ge 1\).

(b) Find \(I_3\).

(c) Use the substitution \(x = \sin\theta\) to explain why \(I_3\) equals \(\displaystyle\int_0^{\pi/2}\cos^{7}\theta\,d\theta\).

(a) \(\dfrac{d}{dx}\big(x(1 - x^{2})^{n}\big) = (1 - x^{2})^{n} - 2nx^{2}(1 - x^{2})^{n-1}\). ✓
Write \(x^{2} = 1 - (1 - x^{2})\): the derivative is \((2n + 1)(1 - x^{2})^{n} - 2n(1 - x^{2})^{n-1}\). ✓
Integrate from 0 to 1: \(\big[x(1 - x^{2})^{n}\big]_0^1 = 0 - 0 = 0\) (as \(n \ge 1\)), so \(0 = (2n + 1)I_n - 2nI_{n-1}\). ✓
(b) \(I_0 = 1\), so \(I_3 = \tfrac{6}{7}\cdot\tfrac{4}{5}\cdot\tfrac{2}{3} = \tfrac{16}{35}\). ✓
(c) \(dx = \cos\theta\,d\theta\), \(1 - x^{2} = \cos^{2}\theta\), and the limits 0, 1 become 0, \(\tfrac{\pi}{2}\): \[I_3 = \int_0^{\pi/2}\cos^{6}\theta\cos\theta\,d\theta = \int_0^{\pi/2}\cos^{7}\theta\,d\theta.\] ✓
This agrees with the cosine formula: \(\tfrac{6}{7}\cdot\tfrac{4}{5}\cdot\tfrac{2}{3} \times 1 = \tfrac{16}{35}\). ✓

Proof

[7 marks]

Given \(I_n = \displaystyle\int_0^{\pi/4}\sec^{n}x\,dx\).

(a) By considering \(\dfrac{d}{dx}\big(\tan x\sec^{n}x\big)\), show that \((n + 1)I_{n+2} = 2^{n/2} + nI_n\) for \(n \ge 0\).

(b) Given that \(I_1 = \ln\big(1 + \sqrt{2}\big)\), find \(I_5\) in exact form.

(a) Product rule, with \(\dfrac{d}{dx}\sec^{n}x = n\sec^{n}x\tan x\): \[\begin{aligned} \frac{d}{dx}\big(\tan x\sec^{n}x\big) &= \sec^{2}x\sec^{n}x + n\tan^{2}x\sec^{n}x. \end{aligned}\] ✓
Write \(\tan^{2}x = \sec^{2}x - 1\): the derivative is \((n + 1)\sec^{n+2}x - n\sec^{n}x\). ✓
Integrate from 0 to \(\tfrac{\pi}{4}\): \(\big[\tan x\sec^{n}x\big]_0^{\pi/4} = 1 \times (\sqrt{2})^{n} - 0 = 2^{n/2}\). ✓
So \(2^{n/2} = (n + 1)I_{n+2} - nI_n\), that is \((n + 1)I_{n+2} = 2^{n/2} + nI_n\). ✓
(b) \(n = 1\): \(2I_3 = \sqrt{2} + \ln\big(1 + \sqrt{2}\big)\), so \(I_3 = \tfrac{1}{2}\big(\sqrt{2} + \ln(1 + \sqrt{2})\big)\). ✓
\(n = 3\): \(4I_5 = 2\sqrt{2} + 3I_3 = 2\sqrt{2} + \tfrac{3}{2}\sqrt{2} + \tfrac{3}{2}\ln\big(1 + \sqrt{2}\big)\). ✓
\(I_5 = \tfrac{1}{8}\big(7\sqrt{2} + 3\ln(1 + \sqrt{2})\big)\) (about 1.568). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.