Question 5 - A-Level Maths

CAIE FP1 2016 June — Question 5 9 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2016
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reduction Formulae
Type	Mixed trigonometric products
Difficulty	Challenging +1.3 This is a standard reduction formula question requiring differentiation of a given expression, integration by parts technique, and recursive calculation. While it involves Further Maths content (reduction formulae), the method is formulaic: differentiate the hint, integrate both sides, apply limits, and use the recurrence relation. The 'hint' structure makes it more accessible than deriving the reduction formula independently, placing it moderately above average difficulty.
Spec	1.07n Stationary points: find maxima, minima using derivatives 8.06a Reduction formulae: establish, use, and evaluate recursively

5 Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \sin ^ { 2 } x \mathrm {~d} x$, for $n \geqslant 0$. By differentiating $\cos ^ { n - 1 } x \sin ^ { 3 } x$ with respect to $x$, prove that $$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 } \quad \text { for } n \geqslant 2$$ Hence find the exact value of $I _ { 4 }$.

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$\frac{d}{dx}(\cos^{n-1}x\sin^3 x) = -(n-1)\cos^{n-2}x\sin^4 x + 3\cos^n x\sin^2 x$	M1A1
$\Rightarrow \left[\cos^{n-1}x\sin^3 x\right]_0^{\frac{1}{2}\pi} = -\int_0^{\frac{1}{2}\pi}(n-1)\cos^{n-2}x\sin^2 x(1-\cos^2 x)\,dx + 3I_n$	M1
$\Rightarrow 0 = -(n-1)I_{n-2} + (n-1)I_n + 3I_n$	M1
$\Rightarrow (n+2)I_n = (n-1)I_{n-2}$ (AG)	A1 [5]
$I_0 = \int_0^{\frac{1}{2}\pi}\sin^2 x\,dx = \frac{1}{2}\int_0^{\frac{1}{2}\pi}(1-\cos 2x)\,dx$	M1
$= \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\frac{1}{2}\pi} = \frac{\pi}{4}$	A1
$I_2 = \frac{1}{4} \times \frac{\pi}{4}$, $I_4 = \frac{1}{2} \times \frac{1}{4} \times \frac{\pi}{4} = \frac{\pi}{32}$	M1A1 [4]

## Question 5:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\cos^{n-1}x\sin^3 x) = -(n-1)\cos^{n-2}x\sin^4 x + 3\cos^n x\sin^2 x$ | M1A1 | |
| $\Rightarrow \left[\cos^{n-1}x\sin^3 x\right]_0^{\frac{1}{2}\pi} = -\int_0^{\frac{1}{2}\pi}(n-1)\cos^{n-2}x\sin^2 x(1-\cos^2 x)\,dx + 3I_n$ | M1 | |
| $\Rightarrow 0 = -(n-1)I_{n-2} + (n-1)I_n + 3I_n$ | M1 | |
| $\Rightarrow (n+2)I_n = (n-1)I_{n-2}$ **(AG)** | A1 [5] | |
| $I_0 = \int_0^{\frac{1}{2}\pi}\sin^2 x\,dx = \frac{1}{2}\int_0^{\frac{1}{2}\pi}(1-\cos 2x)\,dx$ | M1 | |
| $= \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\frac{1}{2}\pi} = \frac{\pi}{4}$ | A1 | |
| $I_2 = \frac{1}{4} \times \frac{\pi}{4}$, $I_4 = \frac{1}{2} \times \frac{1}{4} \times \frac{\pi}{4} = \frac{\pi}{32}$ | M1A1 [4] | |

---

Show LaTeX source

5 Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \sin ^ { 2 } x \mathrm {~d} x$, for $n \geqslant 0$. By differentiating $\cos ^ { n - 1 } x \sin ^ { 3 } x$ with respect to $x$, prove that

$$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 } \quad \text { for } n \geqslant 2$$

Hence find the exact value of $I _ { 4 }$.

\hfill \mbox{\textit{CAIE FP1 2016 Q5 [9]}}

This paper (13 questions)

View full paper

Q1 4 Q2 6 Q3 6 Q4 8 Q5 9 Q6 9 Q7 10 Q8 11 Q9 11 Q10 12 Q11 EITHER Q11 OR Q15