# Question 11:

## Part: Integration by parts (EITHER approach)
$I=\int e^x\sin x\,dx=-e^x\cos x+\int e^x\cos x\,dx$ | M1 |

$=-e^x\cos x+e^x\sin x-\int e^x\sin x\,dx$ | A1, M1 |

$\therefore 2I=e^x(\sin x-\cos x)$

$\therefore\int_0^\pi e^x\sin x\,dx=\left[\dfrac{1}{2}e^x(\sin x-\cos x)\right]_0^\pi$

$=\dfrac{e^\pi}{2}-\left(-\dfrac{1}{2}\right)=\dfrac{1+e^\pi}{2}$ | A1 | (AG) | 4 marks

## Part: Reduction formula
$I_n=\int_0^\pi e^x\sin^n x\,dx$

$=\left[\sin^n x\cdot e^x\right]_0^\pi-\int_0^\pi e^x(n\sin^{n-1}x\cos x)\,dx$ | M1 |

$=\left\{0-\left[n\sin^{n-1}x\cos x\cdot e^x\right]_0^\pi+n\int_0^\pi e^x(\cos^2 x(n-1)\sin^{n-2}x-\sin^{n-1}x\sin x)\,dx\right\}$ | A1 |

$=0+n(n-1)\int_0^\pi e^x\cos^2 x\sin^{n-2}x\,dx-nI_n$ | A1 | (AG)

$=n(n-1)\int_0^\pi e^x(1-\sin^2 x)\sin^{n-2}x\,dx-nI_n$ | M1A1 |

$\therefore (n+1)I_n=n(n-1)I_{n-2}-n(n-1)I_n$

$\therefore (n(n-1)+n+1)I_n=n(n-1)I_{n-2}$

$\therefore (n^2+1)I_n=n(n-1)I_{n-2}$ | A1 | (AG) | 6 marks

## Part: Uses reduction formula to find I₅
$I_5=\dfrac{20}{26}I_3=\dfrac{20}{26}\times\dfrac{6}{10}I_1$ | M1 |

$\Rightarrow I_5=\dfrac{6}{13}\left(\dfrac{1+e^\pi}{2}\right)=\dfrac{3}{13}(1+e^\pi)$ | A1 |

## Part: Mean value
Mean value $=\dfrac{\int_0^\pi e^x\sin^5 x\,dx}{\pi-0}=\dfrac{3}{13\pi}(1+e^\pi)$ | M1A1 | 4 marks

# Question 11:

## Part 1 (Finding m, 7 marks)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & m-1 \end{vmatrix} = -m\mathbf{i} + 4(1-m)\mathbf{j} + 4\mathbf{k}$ | M1A1 | Obtains direction of common perpendicular |
| $\dfrac{\begin{pmatrix}1\\0\\-4\end{pmatrix}\cdot\begin{pmatrix}-m\\4-4m\\4\end{pmatrix}}{\sqrt{m^2 + 16(1-2m+m^2)+16}} = 3$ | M1A1 | Uses result for shortest distance between lines |
| $\Rightarrow 19m^2 - 40m + 4 = 0$ | A1 | Solves equation |
| $\Rightarrow (19m-2)(m-2) = 0$ | M1 | |
| $\Rightarrow m = 2$, since $m$ is an integer. (AG) | A1 | |

## Part 2 (Shortest distance, 3 marks)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{CA} = \begin{pmatrix}1\\0\\-4\end{pmatrix}$ and $\mathbf{CD} = \begin{pmatrix}0\\1\\1\end{pmatrix}$ or $\mathbf{AD} = \begin{pmatrix}-1\\1\\5\end{pmatrix}$ | B1 | Finds relevant vectors |
| $\dfrac{1}{\sqrt{17}}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\0 & 1 & 1\\1 & 0 & -4\end{vmatrix} = \dfrac{1}{\sqrt{17}}\begin{pmatrix}-4\\1\\-1\end{pmatrix}$ | M1 | Use of cross-product |
| $\dfrac{1}{\sqrt{17}}\sqrt{4^2+1^2+1^2} = \sqrt{\dfrac{18}{17}} \quad (=1.03)$ | A1 | Obtains shortest distance |

## Part 3 (Angle between planes, 4 marks)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{BC} = 3\mathbf{i} - \mathbf{j} + 5\mathbf{k}$ | B1 | Finds $2^{nd}$ vector in BCD (CD may already have been found) |
| $\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\3 & -1 & 5\\0 & 1 & 1\end{vmatrix} = -6\mathbf{i}-3\mathbf{j}+3\mathbf{k} \sim 2\mathbf{i}+\mathbf{j}-\mathbf{k}$ | M1 | Finds normal vector to BCD (Normal to ACD already found) |
| $\cos\theta = \dfrac{(4\mathbf{i}-\mathbf{j}+\mathbf{k})\cdot(2\mathbf{i}+\mathbf{j}-\mathbf{k})}{\sqrt{16+1+1}\sqrt{4+1+1}} = \dfrac{6}{\sqrt{18}\sqrt{6}} = \dfrac{1}{\sqrt{3}}$ | M1 | Finds angle between planes = angle between normal vectors |
| $\therefore$ Angle between planes $= \cos^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$ (AG) | A1 | |

## Part 2 Alternative Method (3 marks)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vector from $D$ to any point on $AC$: $\begin{pmatrix}1+t\\-1\\-5-4t\end{pmatrix}$ | (B1) | Or (a) |
| $\begin{pmatrix}1+t\\-1\\-5-4t\end{pmatrix}\cdot\begin{pmatrix}1\\0\\-4\end{pmatrix} = 0 \Rightarrow t = -\dfrac{21}{17}$ | (M1) | Uses orthogonality to obtain $t$ |
| $\dfrac{1}{\sqrt{17}}\sqrt{4^2+1^2+1^2} = \sqrt{\dfrac{18}{17}} \quad (=1.03)$ | (A1) | Finds magnitude of perpendicular |
| $\lvert\overrightarrow{AD}\rvert = \sqrt{27}$ | (B1) | Or (b): Finds length of $AD$ (or $CD$) |
| $\dfrac{\begin{vmatrix}\begin{pmatrix}-1\\1\\5\end{pmatrix}\cdot\begin{pmatrix}1\\0\\-4\end{pmatrix}\end{vmatrix}}{\sqrt{4^2+1^2}} = \dfrac{21}{\sqrt{17}}$ | (M1) | Finds projection of $AD$ (or $CD$) onto $AC$ |
| $\sqrt{27 - \dfrac{441}{17}} = \sqrt{\dfrac{18}{17}} \quad (=1.03)$ | (A1) | Finds perpendicular by Pythagoras |