## Question 13:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = a\sin^2\theta \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2a\sin\theta\cos\theta$ | B1 | $\frac{\mathrm{d}x}{\mathrm{d}\theta}=2a\sin\theta\cos\theta$ seen or implied by substitution |
| $\int x^{\frac{1}{2}}\sqrt{a-x}\,\mathrm{d}x = \int\sqrt{a}\sin\theta\sqrt{a-a\sin^2\theta}\times 2a\sin\theta\cos\theta\,\{\mathrm{d}\theta\}$ | M1 | Attempts to substitute, fully replacing $x^{\frac{1}{2}}$ and $\sqrt{a-x}$ with $\theta$'s and $\mathrm{d}x$ with their $\mathrm{d}x=...$. See detailed bullet-point criteria in notes. |
| $= 2a^2\int\sin^2\theta\cos^2\theta\,\mathrm{d}\theta = 2a^2\int\!\left(\tfrac{1}{2}\sin2\theta\right)^2\mathrm{d}\theta$ | dM1 | Attempts to use $\sin2\theta=2\sin\theta\cos\theta$ to convert integral of form $\int\sin^2\theta\cos^2\theta\,\mathrm{d}\theta$ |
| Replaces/considers limits: $\{x=0\Rightarrow\}\theta=0$, $\{x=a\Rightarrow\}\theta=\frac{\pi}{2}$; gives $= \frac{1}{2}a^2\int_0^{\frac{\pi}{2}}\sin^2 2\theta\,\mathrm{d}\theta$ | A1* | Replaces/considers limits at some stage before given answer, proceeds with no errors to given answer. Must see integral sign with correct limits and $\mathrm{d}\theta$. |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $...\int\sin^2 2\theta\,\mathrm{d}\theta \rightarrow ...\int\frac{1-\cos4\theta}{2}\,\mathrm{d}\theta$ | M1 | Adopts appropriate strategy using double angle identity to obtain integrable form |
| $\rightarrow ...\left(\frac{\theta}{2} - \frac{1}{8}\sin4\theta\right)$ | dM1 | Integrates into form $\pm p\theta \pm q\sin4\theta$ |
| $\mu\int\sin^2 2\theta\,\mathrm{d}\theta \rightarrow \frac{\mu}{2}\!\left(\theta - \frac{1}{4}\sin4\theta\right)$ | A1 | Correct integration of $\mu\int\sin^2 2\theta\,\mathrm{d}\theta$. Here $\mu$ may be 1. Condone lack of limits here. |
| $\left\{\int_0^a x^{\frac{1}{2}}\sqrt{a-x}\,\mathrm{d}x\right\} = \frac{a^2}{4}\!\left(\frac{\pi}{2}-0-0\right) = \frac{1}{8}\pi a^2$ | A1 | Applies limits to correct integral and proceeds to $\frac{1}{8}\pi a^2$ following correct work. Note $\frac{a^2}{4}\!\left[\theta+\frac{1}{4}\sin4\theta\right]_0^{\frac{\pi}{2}}=\frac{1}{8}\pi a^2$ is **incorrect** and scores M1dM1A0A0. |

## Question 13(b) Alternative via IBP Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| IBP with $u = \sin 2\theta$, $v' = \sin 2\theta$, $u' = 2\cos 2\theta$, $v = -\frac{\cos 2\theta}{2}$ | — | Setup shown |
| $= \left[-\frac{\sin 2\theta \cos 2\theta}{2}\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -\cos^2 2\theta \, d\theta$ | — | IBP applied |
| $\Rightarrow 2\int_0^{\frac{\pi}{2}} \sin^2 2\theta \, d\theta = \left[-\frac{\sin 2\theta \cos 2\theta}{2}\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 1 \, d\theta$ | M1 | Score M1 here |
| $\Rightarrow 2\int_0^{\frac{\pi}{2}} \sin^2 2\theta \, d\theta = \left[-\frac{\sin 2\theta \cos 2\theta}{2} + \theta\right]_0^{\frac{\pi}{2}}$ | dM1, A1 | Score dM1 here, A1 if correct including the 2 (ignoring limits) |
| $\Rightarrow \int_0^{\frac{\pi}{2}} \sin^2 2\theta \, d\theta = \frac{1}{2}\left(0 + \frac{\pi}{2}\right) - \frac{1}{2}(0+0) = \frac{\pi}{4}$ | — | Evaluation |
| $\int_0^a x^2\sqrt{a-x} \, dx = \frac{1}{8}\pi a^2$ | A1* | Score A1* with no errors |

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