| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.2 This is a standard Further Maths reduction formula question requiring integration by parts to derive the recurrence relation, then applying it repeatedly to evaluate a definite integral. While it involves multiple steps and careful algebraic manipulation, the techniques are well-practiced in F3 and follow a predictable pattern. Part (c) requires the insight to use cos²x + sin²x = 1 to convert to a form where part (b) applies, adding modest problem-solving beyond routine application. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| GUV SIHI NI JIVM ION OC | VJYV SIHI NI JIIIM ION OC | VJ4V SIHIANI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n=\int\cos^n x\,dx=\int\cos x\cos^{n-1}x\,dx\) | M1 | Correct split; could be implied by their work |
| \(=\sin x\cos^{n-1}x+\int(n-1)\cos^{n-2}x\sin^2 x\,dx\) | dM1 | Obtains \(p\sin x\cos^{n-1}x+\int q\cos^{n-2}x\sin^2 x\,dx\); requires previous M mark |
| \(=\sin x\cos^{n-1}x+\int(n-1)\cos^{n-2}x(1-\cos^2 x)\,dx\) | A1 | Replaces \(\sin^2 x\) with \(1-\cos^2 x\) |
| \(\Rightarrow I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}\) * | A1* | Proceeds to given answer with at least one intermediate step; no errors; clear bracketing errors must be recovered before given answer |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n=\int\cos^n x\,dx=\int\cos^2 x\cos^{n-2}x\,dx=\int(1-\sin^2 x)\cos^{n-2}x\,dx\) | M1 | Correct split and replaces \(\cos^2 x\) with \(1-\sin^2 x\) |
| \(=\int(\cos^{n-2}x-\cos^{n-2}x\sin^2 x)\,dx=\int\cos^{n-2}x\,dx-\int(\sin x\sin x\cos^{n-2}x)\,dx=...\) | dM1 A1 | M1: expands, splits, obtains \(p\int\cos^{n-2}x\,dx+q\cos^{n-1}x\sin x+\int r\cos^n x\,dx\); A1: correct expression for \(I_n\) |
| \(\Rightarrow I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}\) * | A1* | At least one intermediate step, no errors; bracketing errors recovered before given answer |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n=\frac{1}{n}\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+\frac{n-1}{n}I_{n-2}\) or \(=\frac{1}{n}(n-1)I_{n-2}\); \(I_2=\frac{1}{2}\left[\cos^{2-1}x\sin x\right]_0^{\frac{\pi}{2}}+\frac{2-1}{2}I_0\) or \(=\frac{1}{2}I_0\) | M1 | Uses RF to obtain \(I_n\) in terms of \(I_{n-2}\) or \(I_2\) in terms of \(I_0\); condone if limits absent |
| \(I_n=\frac{(n-1)(n-3)...5\times3\times1}{n(n-2)(n-4)...6\times4\times2}I_0\) with dots and at least 3 terms in each product | A1 | Correct expression for \(I_n\) in terms of \(I_0\) following correct work including 2 applications of reduction formula |
| \(I_0=\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{2}\) or \(I_0=[x]_0^{\frac{\pi}{2}}=\frac{\pi}{2}\) or \(I_0=\frac{\pi}{2}-0\) | B1 | Correct value for \(I_0\); requires written evidence of integration |
| \(\therefore I_n=\frac{(n-1)(n-3)...5\times3\times1}{n(n-2)(n-4)...6\times4\times2}\times\frac{\pi}{2}\) * | A1* | Requires all previous marks; withhold if \(\frac{1}{k}\left[\cos^{k-1}x\sin x\right]_0^{\frac{\pi}{2}}\) seen or expression disappears without substitution |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^{\frac{\pi}{2}}\cos^6 x\sin^2 x\,dx=\int_0^{\frac{\pi}{2}}\cos^6 x(1-\cos^2 x)\,dx\) | M1 | Replaces \(\sin^2 x\) with \(1-\cos^2 x\); can be implied by attempt at \(I_6-I_8\) |
| \(=I_6-I_8=\left(\frac{5\times3\times1}{6\times4\times2}-\frac{7\times5\times3\times1}{8\times6\times4\times2}\right)\frac{\pi}{2}\) | A1 | Any correct numerical expression for the integral |
| \(=\frac{5}{32}\pi-\frac{35}{256}\pi=\frac{5}{256}\pi\) | A1 | Correct exact value; accept equivalent fractions e.g. \(\left(\frac{5}{128}\right)\frac{\pi}{2}\) |
| (3) Total 11 | "Hence" question — must show clear use of \(I_6-I_8\); no credit in (b) for work in (c); just "\(I=\frac{5}{256}\pi\)" is 0/3 but "\(I_6-I_8=\frac{5}{256}\pi\)" is 3/3 |
## Question 8(a) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n=\int\cos^n x\,dx=\int\cos x\cos^{n-1}x\,dx$ | M1 | Correct split; could be implied by their work |
| $=\sin x\cos^{n-1}x+\int(n-1)\cos^{n-2}x\sin^2 x\,dx$ | dM1 | Obtains $p\sin x\cos^{n-1}x+\int q\cos^{n-2}x\sin^2 x\,dx$; requires previous M mark |
| $=\sin x\cos^{n-1}x+\int(n-1)\cos^{n-2}x(1-\cos^2 x)\,dx$ | A1 | Replaces $\sin^2 x$ with $1-\cos^2 x$ |
| $\Rightarrow I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}$ * | A1* | Proceeds to given answer with at least one intermediate step; no errors; clear bracketing errors must be recovered before given answer |
| | **(4)** | |
## Question 8(a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n=\int\cos^n x\,dx=\int\cos^2 x\cos^{n-2}x\,dx=\int(1-\sin^2 x)\cos^{n-2}x\,dx$ | M1 | Correct split and replaces $\cos^2 x$ with $1-\sin^2 x$ |
| $=\int(\cos^{n-2}x-\cos^{n-2}x\sin^2 x)\,dx=\int\cos^{n-2}x\,dx-\int(\sin x\sin x\cos^{n-2}x)\,dx=...$ | dM1 A1 | M1: expands, splits, obtains $p\int\cos^{n-2}x\,dx+q\cos^{n-1}x\sin x+\int r\cos^n x\,dx$; A1: correct expression for $I_n$ |
| $\Rightarrow I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}$ * | A1* | At least one intermediate step, no errors; bracketing errors recovered before given answer |
| | **(4)** | |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n=\frac{1}{n}\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+\frac{n-1}{n}I_{n-2}$ or $=\frac{1}{n}(n-1)I_{n-2}$; $I_2=\frac{1}{2}\left[\cos^{2-1}x\sin x\right]_0^{\frac{\pi}{2}}+\frac{2-1}{2}I_0$ or $=\frac{1}{2}I_0$ | M1 | Uses RF to obtain $I_n$ in terms of $I_{n-2}$ or $I_2$ in terms of $I_0$; condone if limits absent |
| $I_n=\frac{(n-1)(n-3)...5\times3\times1}{n(n-2)(n-4)...6\times4\times2}I_0$ with dots and at least 3 terms in each product | A1 | Correct expression for $I_n$ in terms of $I_0$ following correct work including 2 applications of reduction formula |
| $I_0=\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{2}$ or $I_0=[x]_0^{\frac{\pi}{2}}=\frac{\pi}{2}$ or $I_0=\frac{\pi}{2}-0$ | B1 | Correct value for $I_0$; requires written evidence of integration |
| $\therefore I_n=\frac{(n-1)(n-3)...5\times3\times1}{n(n-2)(n-4)...6\times4\times2}\times\frac{\pi}{2}$ * | A1* | Requires all previous marks; withhold if $\frac{1}{k}\left[\cos^{k-1}x\sin x\right]_0^{\frac{\pi}{2}}$ seen or expression disappears without substitution |
| | **(4)** | |
---
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{\pi}{2}}\cos^6 x\sin^2 x\,dx=\int_0^{\frac{\pi}{2}}\cos^6 x(1-\cos^2 x)\,dx$ | M1 | Replaces $\sin^2 x$ with $1-\cos^2 x$; can be implied by attempt at $I_6-I_8$ |
| $=I_6-I_8=\left(\frac{5\times3\times1}{6\times4\times2}-\frac{7\times5\times3\times1}{8\times6\times4\times2}\right)\frac{\pi}{2}$ | A1 | Any correct numerical expression for the integral |
| $=\frac{5}{32}\pi-\frac{35}{256}\pi=\frac{5}{256}\pi$ | A1 | Correct exact value; accept equivalent fractions e.g. $\left(\frac{5}{128}\right)\frac{\pi}{2}$ |
| | **(3)** **Total 11** | "Hence" question — must show clear use of $I_6-I_8$; no credit in (b) for work in (c); just "$I=\frac{5}{256}\pi$" is 0/3 but "$I_6-I_8=\frac{5}{256}\pi$" is 3/3 |8.
$$I _ { n } = \int \cos ^ { n } x \mathrm {~d} x \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that for $n \geqslant 2$
$$I _ { n } = \frac { 1 } { n } \cos ^ { n - 1 } x \sin x + \frac { n - 1 } { n } I _ { n - 2 }$$
\item Show that for positive even integers $n$
$$\int _ { 0 } ^ { \overline { 2 } } \cos ^ { n } x d x = \frac { ( n - 1 ) ( n - 3 ) \ldots 5 \times 3 \times 1 } { n ( n - 2 ) ( n - 4 ) \ldots 6 \times 4 \times 2 } \times \overline { 2 }$$
\item Hence determine the exact value of
$$\int _ { 0 } ^ { \overline { 2 } } \cos ^ { 6 } x \sin ^ { 2 } x d x$$
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GUV SIHI NI JIVM ION OC & VJYV SIHI NI JIIIM ION OC & VJ4V SIHIANI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel F3 2023 Q8 [11]}}