# Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \pm\frac{1}{\sqrt{1-k\sqrt{x}^2}} \times \ldots x^{-\frac{1}{2}}$ or $\cos y = 2x^{\frac{1}{2}} \Rightarrow \pm\sin y \frac{dy}{dx} = \ldots x^{-\frac{1}{2}}$ | M1 | Attempts arccos derivative formula with chain rule; $k$ may be 1 |
| $\frac{dy}{dx} = \pm\frac{1}{\sqrt{1-4x}}\times\!\left(Kx^{-\frac{1}{2}}\right)$ or $\frac{dy}{dx} = \pm\frac{Kx^{-\frac{1}{2}}}{\sqrt{1-(2\sqrt{x})^2}}$ | dM1 | Correct overall form, may have errors in sign/constants with $k\neq1$ |
| $\frac{dy}{dx} = -\frac{1}{\sqrt{x}\sqrt{1-4x}}$ | A1 | Need not be simplified; award when first seen |

# Question 8(b) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int y\,dx = \int 1\times\arccos(2\sqrt{x})\,dx = x\arccos(2\sqrt{x}) - \int x \cdot \frac{-1}{\sqrt{x}\sqrt{1-4x}}\,dx$ | M1 | Attempts integration by parts on $1\times\arccos(2\sqrt{x})$ |
| $= x\arccos(2\sqrt{x}) + \int\frac{\sqrt{x}}{\sqrt{1-4x}}\,dx$* | A1* | Correct; clear IBP statement required before printed answer |

# Question 8(b) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}\!\left(x\arccos(2\sqrt{x})\right) = 1\cdot\arccos(2\sqrt{x}) + x\cdot\frac{-1}{\sqrt{x}\sqrt{1-4x}}$ | M1 | Applies product rule |
| $\Rightarrow \int\arccos(2\sqrt{x})\,dx = x\arccos(2\sqrt{x}) + \int\frac{\sqrt{x}}{\sqrt{1-4x}}\,dx$* | A1* | Rearranges and integrates to reach printed answer, no errors |

# Question 8(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2\sqrt{x}}\frac{dx}{d\theta} = -\frac{1}{2}\sin\theta$, $dx = -\sqrt{x}\sin\theta\,d\theta$, $\frac{dx}{d\theta} = -\frac{1}{2}\sin\theta\cos\theta = -\frac{1}{4}\sin 2\theta$ | B1 | Any correct expression involving $dx$ and $d\theta$ |
| $\int\frac{\sqrt{x}}{\sqrt{1-4x}}\,dx = \int\frac{-(\frac{1}{2}\cos\theta)^2\sin\theta}{\sqrt{1-4(\frac{1}{2}\cos\theta)^2}}\,d\theta$ | M1 | Complete substitution to achieve integral in $\theta$ only |
| $= -\frac{1}{4}\int\frac{\cos^2\theta\sin\theta}{\sqrt{1-\cos^2\theta}}\,d\theta = -\frac{1}{4}\int\cos^2\theta\,d\theta$ | A1 | Correct simplified integral (aside from limits) |
| $x=0\Rightarrow\theta=\frac{\pi}{2}$; $x=\frac{1}{8}\Rightarrow\theta=\frac{\pi}{4}$; so $\int_0^{\frac{1}{8}}\frac{\sqrt{x}}{\sqrt{1-4x}}\,dx = \frac{1}{4}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos^2\theta\,d\theta$ | A1 | Correct limits and sign; accept equivalent limits e.g. $-\frac{\pi}{4}$ to $-\frac{\pi}{4}$ or $\frac{\pi}{2}$ to $\frac{3\pi}{4}$ |

# Question 8(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{4}\int\frac{1}{2}(1+\cos 2\theta)\,d\theta = K\!\left(\theta \pm \frac{1}{2}\sin 2\theta\right)$ | M1 | Uses double angle identity; accept $\cos^2\theta = \frac{1}{2}(\pm1\pm\cos 2\theta)$, look for $1\to\theta$ and $\cos 2\theta\to\pm\frac{1}{2}\sin 2\theta$ |
| $\int_0^{\frac{1}{8}}\frac{\sqrt{x}}{\sqrt{1-4x}}\,dx = \frac{1}{8}\!\left[\theta+\frac{1}{2}\sin 2\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \ldots = \frac{\pi}{32} - \frac{1}{16}$ | dM1 | Applies limits (either way) in $\theta$, or reverse substitution with limits 0 and $\frac{1}{8}$ |
| $\Rightarrow \int_0^{\frac{1}{8}}\arccos(2\sqrt{x})\,dx = \left[x\arccos 2\sqrt{x}\right]_0^{\frac{1}{8}} + \frac{\pi}{32} - \frac{1}{16} = \frac{1}{8}\arccos\frac{1}{\sqrt{2}} - 0 + \frac{\pi}{32} - \frac{1}{16}$ | dM1 | Applies limits 0 and $\frac{1}{8}$ to $x\arccos(2\sqrt{x})$ and combines with other integral result |
| $= \frac{\pi}{16} - \frac{1}{16}$ | A1 | Correct final answer |