## Question 9:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(2x + x) = \sin 2x\cos x + \cos 2x\sin x$ | M1 | Attempts $\sin(A+B)$ with $A=2x, B=x$ or vice versa. Accept $\sin(2x+x) = \sin 2x\cos x \pm \cos 2x\sin x$ |
| $= 2\sin x\cos x\cos x + \left(1 - 2\sin^2 x\right)\sin x$ | dM1 | Uses correct double angle identities for $\sin 2x$ and $\cos 2x$. If $\cos 2x = \cos^2 x - \sin^2 x$ used, then $\cos^2 x$ term must be changed to $1 - \sin^2 x$ later. **Dependent on first M mark** |
| $= 2\sin x\left(1 - \sin^2 x\right) + \left(1 - 2\sin^2 x\right)\sin x$ | M1 | Reaches expression in terms of $\sin x$ only by use of $\cos^2 x = 1 - \sin^2 x$ |
| $= 3\sin x - 4\sin^3 x$ | A1 | $\sin 3x \equiv 3\sin x - 4\sin^3 x$ or $\sin 3x \equiv 3\sin x + -4\sin^3 x$ |

## Question 9(b):

$$\int \sin 3x \cos x \, dx = \int \left(P\sin x \cos x - Q\sin^3 x \cos x\right) dx$$

**AND one of:**

$$\int P\sin x \cos x \, dx = k\sin^2 x \text{ or } k\cos^2 x \text{ or } k\cos 2x$$

**or**

$$\int Q\sin^3 x \cos x \, dx = k\sin^4 x$$

**or**

$$\int Q\sin^3 x \cos x \, dx = \alpha\cos 2x + \beta\cos 4x$$ | M1 | Must use part (a) result; needs correct substitution AND evidence of integrating at least one term correctly

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Examples of correct integration:
- $= \frac{3}{2}\sin^2 x - \sin^4 x (+c)$
- $-\frac{3}{2}\cos^2 x - \sin^4 x (+c)$
- $-\frac{3}{4}\cos 2x - \sin^4 x (+c)$
- $= \frac{3}{2}\sin^2 x + \frac{1}{2}\cos 2x - \frac{1}{8}\cos 4x$
- $-\frac{3}{2}\cos^2 x + \frac{1}{2}\cos 2x - \frac{1}{8}\cos 4x (+c)$
- $-\frac{3}{4}\cos 2x + \frac{1}{2}\cos 2x - \frac{1}{8}\cos 4x (+c)$
- $-\frac{1}{4}\cos 2x - \frac{1}{8}\cos 4x (+c)$ | A1 | Correct integration

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$$\left[\frac{3}{2}\sin^2 x - \sin^4 x\right]_{\pi/6}^{\pi/2} = \frac{3}{2}\sin^2\!\left(\frac{\pi}{2}\right) - \sin^4\!\left(\frac{\pi}{2}\right) - \left\{\frac{3}{2}\sin^2\!\left(\frac{\pi}{6}\right) - \sin^4\!\left(\frac{\pi}{6}\right)\right\}$$ | dM1 | Substitutes both $x = \frac{\pi}{2}$ and $x = \frac{\pi}{6}$ and subtracts either way round. **Dependent upon previous M mark.**

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$$= \frac{1}{2} - \frac{5}{16} = \frac{3}{16}$$ | $\frac{3}{16}$ or $0.1875$ (or exact equivalent) | A1 |

**(4 marks total)**

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### Alternative 1 for 9(b):

$$u = \sin x \Rightarrow \frac{du}{dx} = \cos x \Rightarrow du = \cos x\, dx$$

$$\int \left(P\sin x - Q\sin^3 x\right)\cos x\, dx = \int \left(Pu - Qu^3\right) du$$

**AND one of:**

$$\int Pu\, du = ku^2 \quad \text{or} \quad \int Qu^3\, du = ku^4$$ | M1 | Correct substitution and structure

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$$= \frac{3}{2}u^2 - u^4 (+c)$$ | A1 | Correct integration

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$$\left[\frac{3}{2}u^2 - u^4\right]_{1/2}^{1} = \frac{3}{2} - 1 - \left(\frac{3}{8} - \frac{1}{16}\right)$$

Substitutes both $u=1$ and $u=\frac{1}{2}$, or replaces $u$ with $\sin x$ and substitutes $x=\frac{\pi}{2}$ and $x=\frac{\pi}{6}$ | dM1 | Dependent upon previous M mark

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$$= \frac{1}{2} - \frac{5}{16} = \frac{3}{16}$$ | $\frac{3}{16}$ or $0.1875$ | A1 |

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### Alternative 2 for 9(b):

Integration by parts in correct direction **AND one of** the same integrals as main scheme | M1 |

$$= 3\sin^2 x - 4\sin^4 x - \frac{3}{2}\sin^2 x + 3\sin^4 x (+c)$$ | A1 | Correct integration

Substitutes both $x=\frac{\pi}{2}$ and $x=\frac{\pi}{6}$ and subtracts | dM1 | Dependent upon previous M mark

$$= \frac{1}{2} - \frac{5}{16} = \frac{3}{16}$$ | $\frac{3}{16}$ or $0.1875$ | A1 |

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