| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.8 Part (a) is a standard reduction formula derivation using integration by parts, which is textbook material for FP3. Part (b) is more challenging, requiring recognition that the integral can be solved by substitution (u = sin x) rather than direct application of the reduction formula, then evaluating a polynomial integral. The combination of a routine proof with a non-trivial application that requires insight beyond mechanical formula application places this above average difficulty. |
| Spec | 1.08i Integration by parts4.08f Integrate using partial fractions |
| Answer | Marks |
|---|---|
| \(I_n = \int_0^{\frac{\pi}{2}} \sin x \cdot \sin^{n-1}x \, dx\) | |
| \(= \left[-\cos x \sin^{n-1}x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \cdot (n-1)\sin^{n-2}x \cos x \, dx\) | M1 A1 |
| \(= 0 + \ldots\) | A1 |
| \(= (n-1)\int_0^{\frac{\pi}{2}} \sin^{n-2}x(1-\sin^2 x) dx\) | |
| \(= (n-1)I_{n-2} - (n-1)I_n\) | M1 |
| Leading to \(I_n = \frac{n-1}{n}I_{n-2}\) ⭐ | A1 (5) |
| Answer | Marks |
|---|---|
| \(\int_0^{\frac{\pi}{2}} x(\sin^5 x \cos x) dx = \left[\frac{x\sin^6 x}{6}\right]_0^{\frac{\pi}{2}} - \frac{1}{6}\int_0^{\frac{\pi}{2}} \sin^6 x \, dx\) | M1 A1 |
| \(I_6 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2}I_0 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\left(-\frac{5\pi}{32}\right)\) | M1 A1 |
| Hence \(\int_0^{\frac{\pi}{2}} x(\sin^5 x \cos x) dx = \frac{\pi}{12} - \frac{1}{6} \times \frac{5\pi}{32} = \frac{11\pi}{192}\) | A1 (5) |
**(a)**
$I_n = \int_0^{\frac{\pi}{2}} \sin x \cdot \sin^{n-1}x \, dx$ | |
$= \left[-\cos x \sin^{n-1}x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \cdot (n-1)\sin^{n-2}x \cos x \, dx$ | M1 A1 |
$= 0 + \ldots$ | A1 |
$= (n-1)\int_0^{\frac{\pi}{2}} \sin^{n-2}x(1-\sin^2 x) dx$ | |
$= (n-1)I_{n-2} - (n-1)I_n$ | M1 |
Leading to $I_n = \frac{n-1}{n}I_{n-2}$ ⭐ | A1 (5) |
**(b)**
$\int_0^{\frac{\pi}{2}} x(\sin^5 x \cos x) dx = \left[\frac{x\sin^6 x}{6}\right]_0^{\frac{\pi}{2}} - \frac{1}{6}\int_0^{\frac{\pi}{2}} \sin^6 x \, dx$ | M1 A1 |
$I_6 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2}I_0 = \frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\left(-\frac{5\pi}{32}\right)$ | M1 A1 |
Hence $\int_0^{\frac{\pi}{2}} x(\sin^5 x \cos x) dx = \frac{\pi}{12} - \frac{1}{6} \times \frac{5\pi}{32} = \frac{11\pi}{192}$ | A1 (5) |
(10 marks)5.
$$\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { \mathrm { n } } x \mathrm { dx } , \mathrm { n } \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that $I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }$, for $n \geqslant 2$
\item Using the result in part (a), find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \sin ^ { 5 } x \cos x d x$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 Q5 [10]}}