## Question 8:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(x) = \dfrac{3}{13 + 6 \times \dfrac{2t}{1+t^2} - 5 \times \dfrac{1-t^2}{1+t^2}}$ | M1 | Uses one correct substitution |
| $= \dfrac{3(1+t^2)}{13(1+t^2)+12t-5(1-t^2)}$ | M1 | Both substitutions correct and multiplies numerator and denominator by $1+t^2$ |
| $= \dfrac{3(1+t^2)}{18t^2+12t+8}$, e.g. $\dfrac{3(1+t^2)}{2(9t^2+6t+1)+6}$ or $\dfrac{3(1+t^2)}{2[(3t+1)^2-1]+8}$ | A1* | Must show intermediate step simplifying denominator |
| $\Rightarrow \dfrac{3(1+t^2)}{2(3t+1)^2+6}$ | | Completes to correct expression with no errors |

**(3 marks)**

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### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(x)=\dfrac{3}{7} \Rightarrow \dfrac{3(1+t^2)}{2(3t+1)^2+6}=\dfrac{3}{7} \Rightarrow 21+21t^2=54t^2+36t+24$ | M1 | Equates result from (a) to $\dfrac{3}{7}$ and simplifies to a 3TQ |
| $\Rightarrow 11t^2+12t+1=0$ | | |
| $\Rightarrow (11t+1)(t+1)=0 \Rightarrow t = \ldots$ | M1 | Solves equation by any valid means |
| $t=-1,\ t=-\dfrac{1}{11}$ | A1 | Correct values for $t$ |
| $\Rightarrow x = 2\arctan(\text{"their } t\text{"}) + 2\pi$ for a negative $t$ | dM1 | Dependent on first M1; applies correct process for at least one $x$ from negative $t$. If two positive values found in error, mark cannot be scored |
| $x = \dfrac{3\pi}{2}$ or awrt 4.71 and awrt $x=6.10$ | A1 | Both answers correct and no others in range |

**(5 marks)**

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### Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int f(x)\,dx = \int \dfrac{3(1+t^2)}{2(3t+1)^2+6} \times \dfrac{2}{1+t^2}\,dt = \int \dfrac{3}{(3t+1)^2+3}\,dt$ | B1 | Applies substitution including use of $dx = \dfrac{2}{1+t^2}\,dt$ |
| $= K\arctan(M(3t+1))$ or with $u=(3t+1) \Rightarrow K\arctan(Mu)$ | M1 | Attempts integration to achieve $K\arctan(M(1+3t))$ or $K\arctan(Mu)$ |
| $= \dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{3t+1}{\sqrt{3}}\right)$ | A1 | Correct integral |
| $\displaystyle\int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} f(x)\,dx = \int_{\frac{\pi}{3}}^{\pi} f(x)\,dx + \int_{\pi}^{\frac{4\pi}{3}} f(x)\,dx$ | B1 | Changes limits and splits the integral around $\pi$ |
| $= \int_{\frac{\sqrt{3}}{3}}^{\infty}\ldots\,dt + \int_{-\infty}^{-\sqrt{3}}\ldots\,dt$ or equivalent in $u$ | | |
| $= \dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{3(-\sqrt{3})+1}{\sqrt{3}}\right) - \dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{3\!\left(\dfrac{\sqrt{3}}{3}\right)+1}{\sqrt{3}}\right)+\ldots$ | M1 | Applies limits $\frac{1}{\sqrt{3}}$ and $-\sqrt{3}$ to integrand |
| $= \dfrac{\sqrt{3}}{3}\!\left(\arctan\!\left(\dfrac{\sqrt{3}-9}{3}\right) - \arctan\!\left(\dfrac{\sqrt{3}+3}{3}\right)\right)+\ldots$ | A1 | Correct arctan expressions |
| $= \ldots + \lim_{t\to\infty}\dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{3t+1}{\sqrt{3}}\right) - \lim_{t\to-\infty}\dfrac{1}{\sqrt{3}}\arctan\!\left(\dfrac{3t+1}{\sqrt{3}}\right) = \ldots + \dfrac{\pi}{2\sqrt{3}} - \left(-\dfrac{\pi}{2\sqrt{3}}\right)$ | M1 | Correct work to evaluate the $\pm\infty$ limits |
| $= \dfrac{\sqrt{3}}{3}\!\left(\arctan\!\left(\dfrac{\sqrt{3}-9}{3}\right) - \arctan\!\left(\dfrac{\sqrt{3}+3}{3}\right) + \pi\right)$ | A1 | Fully correct solution |

**(8 marks)**

**Total: 16 marks**