# Question 9:

## Part (a)

| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_n = \int \sin^n 2x\, dx = \int \sin^{n-1} 2x \sin 2x\, dx$; leading to $I_n = [\lambda \sin^{n-1} 2x \cos 2x] - \mu \int \sin^{n-2} 2x \cos^2 2x\, dx$ | M1 | 2.1 |
| $I_n = \left[-\frac{1}{2}\sin^{n-1} 2x \cos 2x\right] + \int (n-1)\sin^{n-2} 2x \cos^2 2x\, dx$ | A1 | 1.1b |
| $I_n = 0 + (n-1)\int \sin^{n-2} 2x(1 - \sin^2 2x)\, dx = (n-1)\int \sin^{n-2} 2x\, dx - (n-1)\int \sin^n 2x\, dx$ | dM1 | 1.1b |
| $nI_n = (n-1)I_{n-2}$ þ $I_n = \frac{n-1}{n}I_{n-2}$ | A1* | 2.1 |

**Alternative:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_n = \int \sin^{n-2} 2x(1 - \cos^2 2x)\, dx = \int \sin^{n-2} 2x\, dx - \int \sin^{n-2} 2x \cos^2 2x\, dx$ | M1 | 1.1b |
| $I_n = I_{n-2} - \int (\sin^{n-2} 2x \cos 2x)(\cos 2x)\, dx$; leads to attempt at integration by parts: $I_n = I_{n-2} - \left[\frac{1}{2(n-1)}\sin^{n-1} 2x \cos 2x\right] - \frac{1}{n-1}\int \sin^n 2x\, dx$ | dM1 | 2.1 |
| $I_n = I_{n-2} - \left[\frac{1}{2(n-1)}\sin^{n-1} 2x \cos 2x\right] - \frac{1}{n-1}I_n$ | A1 | 1.1b |
| $I_n = I_{n-2} - \frac{1}{n-1}I_n \Rightarrow (n-1)I_n = (n-1)I_{n-2} - I_n \Rightarrow I_n = \frac{n-1}{n}I_{n-2}$ | A1* | 2.1 |

## Part (b)

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_0^{\pi/2} 64\sin^5 x \cos^5 x\, dx = \int_0^{\pi/2} A\sin^5 2x\, dx$; Note $A = 2$ | M1 | 2.1 |
| $I_5 = \frac{4}{5}I_3$, $I_3 = \frac{2}{3}I_1$ and $I_1 = \int_0^{\pi/2} \sin 2x\, dx = [\alpha \cos 2x]_0^{\pi/2} = \ldots$ | M1 | 1.1b |
| $= 2 \times \frac{4}{5} \times \frac{2}{3} \times 1 = \frac{16}{15}$ | A1 | 1.1b |

# Question on Reduction Formula (Part a & b):

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\sin^n 2x$ as $\sin^{n-1} 2x \cdot \sin 2x$ and integrates using by parts to the form $I_n = \left[\lambda \sin^{n-1} 2x \cos 2x\right] - \mu\int \sin^{n-2} 2x \cos^2 2x \, dx$ | M1 | |
| Correct integration, may be unsimplified | A1 | |
| Substitutes limits 0 and $\frac{\pi}{2}$ into $uv$; replaces $\cos^2 2x = 1 - \sin^2 2x$ and multiplies out into separate integrals | dM1 | |
| Achieves the printed answer following a correct intermediate line with no errors | A1* | cso |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\sin^n 2x$ as $\sin^{n-2} 2x \cdot \sin^2 2x = \sin^{n-2} 2x(1-\cos^2 2x)$, giving $\int \sin^{n-2} 2x \, dx - \int \sin^{n-2} 2x \cos^2 2x \, dx$ and attempts to integrate | M1 | |
| Integrates by parts to form $I_n = I_{n-2} - \left[\lambda \sin^{n-1} 2x \cos 2x\right] + \mu\int \sin 2x \sin^{n-1} 2x \, dx$ | dM1 | |
| Correct integration, may be unsimplified | A1 | |
| Achieves printed answer following correct intermediate line, no errors | A1* | cso |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses identity $\sin 2x = 2\sin x \cos x$ to write integral as $\int A\sin^5 2x \, dx$ | M1 | |
| Uses answer to part (a) to find $I_5$ and $I_3$; finds $I_1 = \int_0^{\frac{\pi}{2}} \sin 2x \, dx = \ldots$ | M1 | |
| $\dfrac{16}{15}$ | A1 | cso |

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