| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Exponential times trigonometric power |
| Difficulty | Challenging +1.8 This is a challenging Further Maths reduction formula question requiring integration by parts twice, careful algebraic manipulation to derive the recurrence relation, then applying it iteratively with definite integral evaluation. While the technique is standard for F3, the exponential-trigonometric combination creates substantial algebraic complexity and requires multiple applications of the formula, placing it well above average difficulty but within reach of well-prepared FM students. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int e^x \sin^n x\, dx = e^x \sin^n x - n\int e^x \sin^{n-1}x \cos x\, dx\) | M1 | Applies integration by parts to obtain \(\pm e^x \sin^n x \pm \alpha \int e^x \sin^{n-1}x \cos x\, dx\) |
| \(= e^x \sin^n x - n\left\{e^x \sin^{n-1}x \cos x - \int e^x\left((n-1)\sin^{n-2}x\cos^2 x - \sin^n x\right)dx\right\}\) | dM1A1 | M1: Applies integration by parts to \(\pm\alpha\int e^x \sin^{n-1}x\cos x\, dx\) to obtain \(\pm e^x\sin^{n-1}x\cos x \pm \int e^x\left(\alpha\sin^{n-2}x\cos^2 x - \beta\sin^n x\right)dx\). A1: Fully correct expression for \(I_n\) from parts applied twice. |
| Applies \(\cos^2 x = 1 - \sin^2 x\) | dM1 | |
| \(= e^x\sin^n x - n\left\{e^x\sin^{n-1}x\cos x - \int e^x\left((n-1)\sin^{n-2}x - n\sin^n x\right)dx\right\}\) | dM1 | Completes by introducing \(I_{n-2}\) and \(I_n\) and makes \(I_n\) the subject |
| \(I_n = \dfrac{e^x\sin^{n-1}x}{n^2+1}(\sin x - n\cos x) + \dfrac{n(n-1)}{n^2+1}I_{n-2}\) | A1* | Fully correct proof with no errors (but allow e.g. occasional missing "\(dx\)" but any clear errors must be recovered before final answer e.g. missing brackets) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_4 = \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12}{17}I_2\) or \(I_2 = \frac{e^x\sin x}{5}(\sin x - 2\cos x) + \frac{2}{5}I_0\) | M1 | Applies the reduction formula once |
| \(= \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12}{17}\left(\frac{e^x\sin x}{5}(\sin x - 2\cos x) + \frac{2}{5}I_0\right)\) \(= \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12e^x\sin x}{85}(\sin x - 2\cos x) + \frac{24}{85}e^x\) | M1 | Applies the reduction formula again and uses \(I_0 = \int e^x\, dx = e^x\) to obtain an expression in terms of \(x\) |
| \(\int_0^{\frac{\pi}{2}} e^x\sin^4 x\, dx = \left[\frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12e^x\sin x}{85}(\sin x - 2\cos x) + \frac{24}{85}e^x\right]_0^{\frac{\pi}{2}}\) \(= \frac{e^{\frac{\pi}{2}}}{17} + \frac{12e^{\frac{\pi}{2}}}{85} + \frac{24e^{\frac{\pi}{2}}}{85} - \frac{24}{85}\) | dM1 | Uses the limits 0 and \(\frac{\pi}{2}\) and subtracts. Depends on both previous marks. |
| \(= \dfrac{41e^{\frac{\pi}{2}}}{85} - \dfrac{24}{85}\) | A1 | Correct expression or correct values e.g. \(A = \ldots,\ B = \ldots\) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^x \sin^n x\, dx = e^x \sin^n x - n\int e^x \sin^{n-1}x \cos x\, dx$ | M1 | Applies integration by parts to obtain $\pm e^x \sin^n x \pm \alpha \int e^x \sin^{n-1}x \cos x\, dx$ |
| $= e^x \sin^n x - n\left\{e^x \sin^{n-1}x \cos x - \int e^x\left((n-1)\sin^{n-2}x\cos^2 x - \sin^n x\right)dx\right\}$ | dM1A1 | M1: Applies integration by parts to $\pm\alpha\int e^x \sin^{n-1}x\cos x\, dx$ to obtain $\pm e^x\sin^{n-1}x\cos x \pm \int e^x\left(\alpha\sin^{n-2}x\cos^2 x - \beta\sin^n x\right)dx$. A1: Fully correct expression for $I_n$ from parts applied twice. |
| Applies $\cos^2 x = 1 - \sin^2 x$ | dM1 | |
| $= e^x\sin^n x - n\left\{e^x\sin^{n-1}x\cos x - \int e^x\left((n-1)\sin^{n-2}x - n\sin^n x\right)dx\right\}$ | dM1 | Completes by introducing $I_{n-2}$ and $I_n$ and makes $I_n$ the subject |
| $I_n = \dfrac{e^x\sin^{n-1}x}{n^2+1}(\sin x - n\cos x) + \dfrac{n(n-1)}{n^2+1}I_{n-2}$ | A1* | Fully correct proof with no errors (but allow e.g. occasional missing "$dx$" but any clear errors must be recovered before final answer e.g. missing brackets) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_4 = \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12}{17}I_2$ or $I_2 = \frac{e^x\sin x}{5}(\sin x - 2\cos x) + \frac{2}{5}I_0$ | M1 | Applies the reduction formula once |
| $= \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12}{17}\left(\frac{e^x\sin x}{5}(\sin x - 2\cos x) + \frac{2}{5}I_0\right)$ $= \frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12e^x\sin x}{85}(\sin x - 2\cos x) + \frac{24}{85}e^x$ | M1 | Applies the reduction formula again and uses $I_0 = \int e^x\, dx = e^x$ to obtain an expression in terms of $x$ |
| $\int_0^{\frac{\pi}{2}} e^x\sin^4 x\, dx = \left[\frac{e^x\sin^3 x}{17}(\sin x - 4\cos x) + \frac{12e^x\sin x}{85}(\sin x - 2\cos x) + \frac{24}{85}e^x\right]_0^{\frac{\pi}{2}}$ $= \frac{e^{\frac{\pi}{2}}}{17} + \frac{12e^{\frac{\pi}{2}}}{85} + \frac{24e^{\frac{\pi}{2}}}{85} - \frac{24}{85}$ | dM1 | Uses the limits 0 and $\frac{\pi}{2}$ and subtracts. Depends on both previous marks. |
| $= \dfrac{41e^{\frac{\pi}{2}}}{85} - \dfrac{24}{85}$ | A1 | Correct expression or correct values e.g. $A = \ldots,\ B = \ldots$ |6.
$$I _ { n } = \int \mathrm { e } ^ { x } \sin ^ { n } x \mathrm {~d} x \quad n \in \mathbb { Z } \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$I _ { n } = \frac { \mathrm { e } ^ { x } \sin ^ { n - 1 } x } { n ^ { 2 } + 1 } ( \sin x - n \cos x ) + \frac { n ( n - 1 ) } { n ^ { 2 } + 1 } I _ { n - 2 } \quad n \geqslant 2$$
\item Hence find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } e ^ { x } \sin ^ { 4 } x d x$$
giving your answer in the form $A \mathrm { e } ^ { \frac { \pi } { 2 } } + B$ where $A$ and $B$ are rational numbers to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2022 Q6 [10]}}