| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.8 This is a multi-part parametric area question requiring: (a) solving sec³θ = 8 for exact value; (b) deriving area integral using A = ∫y dx/dθ dθ with product rule on x = 3θsinθ; (c) integrating θsec²θ + tanθsec²θ using integration by parts. While systematic, it demands careful algebraic manipulation, exact trigonometric values, and non-trivial integration techniques beyond routine C4 questions. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(y=8\): \(8 = \sec^3\theta \Rightarrow \cos^3\theta = \frac{1}{8} \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\) | M1 | Sets \(y=8\) to find \(\theta\) and attempts to substitute their \(\theta\) into \(x = 3\theta\sin\theta\) |
| \(k = 3\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3}\right)\) so \(k = \frac{\sqrt{3}\,\pi}{2}\) | A1 | Accept \(\frac{\sqrt{3}\,\pi}{2}\) or \(\frac{3\pi}{2\sqrt{3}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = 3\sin\theta + 3\theta\cos\theta\) | B1 | \(3\theta\sin\theta \to 3\sin\theta + 3\theta\cos\theta\); can be implied by later working |
| \(\left\{\int y\frac{dx}{d\theta}\{d\theta\}\right\} = \int(\sec^3\theta)(3\sin\theta + 3\theta\cos\theta)\{d\theta\}\) | M1 | Applies \((\pm K\sec^3\theta)\left(\text{their } \frac{dx}{d\theta}\right)\); ignore integral sign and \(d\theta\); \(K \neq 0\) |
| \(= 3\int \theta\sec^2\theta + \tan\theta\sec^2\theta\, d\theta\) | A1* | Achieves correct result with no errors; must have integral sign and \(d\theta\) in final answer |
| \(x=0\) and \(x=k \Rightarrow \alpha=0\) and \(\beta=\frac{\pi}{3}\) | B1 | \(\alpha=0\) and \(\beta=\frac{\pi}{3}\), or evidence of \(0\to 0\) and \(k\to\frac{\pi}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{\int\theta\sec^2\theta\,d\theta\right\} = \theta\tan\theta - \int\tan\theta\{d\theta\}\) | M1 | \(\theta\sec^2\theta \to A\theta g(\theta) - B\int g(\theta)\), \(A>0\), \(B>0\), where \(g(\theta)\) is a trig function and \(g(\theta) =\) their \(\int\sec^2\theta\,d\theta\). Note: \(g(\theta)\neq\sec^2\theta\) |
| dM1 | Dependent on previous M mark. Either \(\lambda\theta\sec^2\theta \to A\theta\tan\theta - B\int\tan\theta\), \(A>0\), \(B>0\), or \(\theta\sec^2\theta \to \theta\tan\theta - \int\tan\theta\) | |
| \(= \theta\tan\theta - \ln(\sec\theta)\) or \(= \theta\tan\theta + \ln(\cos\theta)\) | A1 | \(\theta\sec^2\theta \to \theta\tan\theta - \ln(\sec\theta)\) or \(\theta\tan\theta + \ln(\cos\theta)\) or with \(\lambda\) factor accordingly |
| \(\left\{\int\tan\theta\sec^2\theta\,d\theta\right\} = \frac{1}{2}\tan^2\theta\) or \(\frac{1}{2}\sec^2\theta\) or \(\frac{1}{2u^2}\) where \(u=\cos\theta\) or \(\frac{1}{2}u^2\) where \(u=\tan\theta\) | M1 | \(\tan\theta\sec^2\theta\) or \(\lambda\tan\theta\sec^2\theta \to \pm C\tan^2\theta\) or \(\pm C\sec^2\theta\) or \(\pm Cu^{-2}\) where \(u=\cos\theta\) |
| A1 | \(\tan\theta\sec^2\theta \to \frac{1}{2}\tan^2\theta\) or \(\frac{1}{2}\sec^2\theta\) or \(\frac{1}{2\cos^2\theta}\) or \(\tan^2\theta - \frac{1}{2}\sec^2\theta\) etc. | |
| \(\{\text{Area}(R)\} = \left[3\theta\tan\theta - 3\ln(\sec\theta) + \frac{3}{2}\tan^2\theta\right]_0^{\frac{\pi}{3}}\) | ||
| \(= \left(3\cdot\frac{\pi}{3}\cdot\sqrt{3} - 3\ln 2 + \frac{3}{2}(3)\right) - (0)\) | ||
| \(= \frac{9}{2} + \sqrt{3}\,\pi - 3\ln 2\) or \(\frac{9}{2} + \sqrt{3}\,\pi + 3\ln\left(\frac{1}{2}\right)\) or \(\frac{9}{2}+\sqrt{3}\,\pi - \ln 8\) or \(\ln\!\left(\frac{1}{8}e^{\frac{9}{2}+\sqrt{3}\pi}\right)\) | A1 o.e. | A decimal answer of 7.861956551… without a correct exact answer is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u=\theta+\tan\theta \Rightarrow \frac{du}{d\theta}=1+\sec^2\theta\); \(\frac{dv}{d\theta}=\sec^2\theta \Rightarrow v=\tan\theta\) | ||
| \(A(\theta+\tan\theta)g(\theta) - B\int(1+h(\theta))g(\theta)\), \(A>0\), \(B>0\) | M1 | |
| \(= (\theta+\tan\theta)\tan\theta - \int(1+\sec^2\theta)\tan\theta\{d\theta\}\) | dM1 | Dependent on previous M mark |
| \(= (\theta+\tan\theta)\tan\theta - \ln(\sec\theta) - \int\tan\theta\sec^2\theta\{d\theta\}\) | A1 | \((\theta+\tan\theta)\tan\theta - \ln(\sec\theta)\) o.e. |
| \(= (\theta+\tan\theta)\tan\theta - \ln(\sec\theta) - \frac{1}{2}\tan^2\theta\) or \(-\frac{1}{2}\sec^2\theta\) etc. | M1, A1 | \(\tan\theta\sec^2\theta \to \pm C\tan^2\theta\) or \(\pm C\sec^2\theta\); then \(\to \frac{1}{2}\tan^2\theta\) or \(\frac{1}{2}\sec^2\theta\) |
| Answer | Marks |
|---|---|
| Note | |
| \(\theta\sec^2\theta \to \theta\tan\theta + \ln(\sec\theta)\) WITH NO INTERMEDIATE WORKING is M0M0A0 | |
| \(\theta\sec^2\theta \to \theta\tan\theta - \ln(\cos\theta)\) WITH NO INTERMEDIATE WORKING is M0M0A0 | |
| \(\theta\sec^2\theta \to \theta\tan\theta - \ln(\sec\theta)\) WITH NO INTERMEDIATE WORKING is M1M1A1 | |
| \(\theta\sec^2\theta \to \theta\tan\theta + \ln(\cos\theta)\) WITH NO INTERMEDIATE WORKING is M1M1A1 | |
| Alternative for \(\int\tan\theta\sec^2\theta\,d\theta\): using \(u=\tan\theta\), \(\frac{du}{d\theta}=\sec^2\theta\), \(\frac{dv}{d\theta}=\sec^2\theta\), \(v=\tan\theta\) gives \(\int\tan\theta\sec^2\theta\,d\theta = \tan^2\theta - \int\tan\theta\sec^2\theta\,d\theta \Rightarrow 2\int\tan\theta\sec^2\theta\,d\theta = \tan^2\theta \Rightarrow \int\tan\theta\sec^2\theta\,d\theta = \frac{1}{2}\tan^2\theta\) | M1, A1 |
| Or using \(u=\sec\theta\): \(\int\tan\theta\sec^2\theta\,d\theta = \sec^2\theta - \int\sec^2\theta\tan\theta\,d\theta \Rightarrow \int\tan\theta\sec^2\theta\,d\theta = \frac{1}{2}\sec^2\theta\) | M1, A1 |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $y=8$: $8 = \sec^3\theta \Rightarrow \cos^3\theta = \frac{1}{8} \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$ | M1 | Sets $y=8$ to find $\theta$ **and** attempts to substitute their $\theta$ into $x = 3\theta\sin\theta$ |
| $k = 3\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3}\right)$ so $k = \frac{\sqrt{3}\,\pi}{2}$ | A1 | Accept $\frac{\sqrt{3}\,\pi}{2}$ or $\frac{3\pi}{2\sqrt{3}}$ |
**Note:** Obtaining two values for $k$ without accepting the correct value is final A0. **[2]**
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = 3\sin\theta + 3\theta\cos\theta$ | B1 | $3\theta\sin\theta \to 3\sin\theta + 3\theta\cos\theta$; can be implied by later working |
| $\left\{\int y\frac{dx}{d\theta}\{d\theta\}\right\} = \int(\sec^3\theta)(3\sin\theta + 3\theta\cos\theta)\{d\theta\}$ | M1 | Applies $(\pm K\sec^3\theta)\left(\text{their } \frac{dx}{d\theta}\right)$; ignore integral sign and $d\theta$; $K \neq 0$ |
| $= 3\int \theta\sec^2\theta + \tan\theta\sec^2\theta\, d\theta$ | A1* | Achieves correct result with no errors; must have integral sign and $d\theta$ in final answer |
| $x=0$ and $x=k \Rightarrow \alpha=0$ and $\beta=\frac{\pi}{3}$ | B1 | $\alpha=0$ and $\beta=\frac{\pi}{3}$, or evidence of $0\to 0$ and $k\to\frac{\pi}{3}$ |
**Note:** Work for the final B1 must be seen in part (b) only. **[4]**
---
## Part (c) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\int\theta\sec^2\theta\,d\theta\right\} = \theta\tan\theta - \int\tan\theta\{d\theta\}$ | M1 | $\theta\sec^2\theta \to A\theta g(\theta) - B\int g(\theta)$, $A>0$, $B>0$, where $g(\theta)$ is a trig function and $g(\theta) =$ their $\int\sec^2\theta\,d\theta$. Note: $g(\theta)\neq\sec^2\theta$ |
| | dM1 | Dependent on previous M mark. Either $\lambda\theta\sec^2\theta \to A\theta\tan\theta - B\int\tan\theta$, $A>0$, $B>0$, **or** $\theta\sec^2\theta \to \theta\tan\theta - \int\tan\theta$ |
| $= \theta\tan\theta - \ln(\sec\theta)$ **or** $= \theta\tan\theta + \ln(\cos\theta)$ | A1 | $\theta\sec^2\theta \to \theta\tan\theta - \ln(\sec\theta)$ or $\theta\tan\theta + \ln(\cos\theta)$ or with $\lambda$ factor accordingly |
| $\left\{\int\tan\theta\sec^2\theta\,d\theta\right\} = \frac{1}{2}\tan^2\theta$ or $\frac{1}{2}\sec^2\theta$ or $\frac{1}{2u^2}$ where $u=\cos\theta$ or $\frac{1}{2}u^2$ where $u=\tan\theta$ | M1 | $\tan\theta\sec^2\theta$ or $\lambda\tan\theta\sec^2\theta \to \pm C\tan^2\theta$ or $\pm C\sec^2\theta$ or $\pm Cu^{-2}$ where $u=\cos\theta$ |
| | A1 | $\tan\theta\sec^2\theta \to \frac{1}{2}\tan^2\theta$ or $\frac{1}{2}\sec^2\theta$ or $\frac{1}{2\cos^2\theta}$ or $\tan^2\theta - \frac{1}{2}\sec^2\theta$ etc. |
| $\{\text{Area}(R)\} = \left[3\theta\tan\theta - 3\ln(\sec\theta) + \frac{3}{2}\tan^2\theta\right]_0^{\frac{\pi}{3}}$ | | |
| $= \left(3\cdot\frac{\pi}{3}\cdot\sqrt{3} - 3\ln 2 + \frac{3}{2}(3)\right) - (0)$ | | |
| $= \frac{9}{2} + \sqrt{3}\,\pi - 3\ln 2$ or $\frac{9}{2} + \sqrt{3}\,\pi + 3\ln\left(\frac{1}{2}\right)$ or $\frac{9}{2}+\sqrt{3}\,\pi - \ln 8$ or $\ln\!\left(\frac{1}{8}e^{\frac{9}{2}+\sqrt{3}\pi}\right)$ | A1 o.e. | A decimal answer of 7.861956551… without a correct exact answer is A0 |
**[6]** Total: **[12]**
---
## Part (c) — Way 2 (Integration by parts on $\int(\theta+\tan\theta)\sec^2\theta\,d\theta$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=\theta+\tan\theta \Rightarrow \frac{du}{d\theta}=1+\sec^2\theta$; $\frac{dv}{d\theta}=\sec^2\theta \Rightarrow v=\tan\theta$ | | |
| $A(\theta+\tan\theta)g(\theta) - B\int(1+h(\theta))g(\theta)$, $A>0$, $B>0$ | M1 | |
| $= (\theta+\tan\theta)\tan\theta - \int(1+\sec^2\theta)\tan\theta\{d\theta\}$ | dM1 | Dependent on previous M mark |
| $= (\theta+\tan\theta)\tan\theta - \ln(\sec\theta) - \int\tan\theta\sec^2\theta\{d\theta\}$ | A1 | $(\theta+\tan\theta)\tan\theta - \ln(\sec\theta)$ o.e. |
| $= (\theta+\tan\theta)\tan\theta - \ln(\sec\theta) - \frac{1}{2}\tan^2\theta$ or $-\frac{1}{2}\sec^2\theta$ etc. | M1, A1 | $\tan\theta\sec^2\theta \to \pm C\tan^2\theta$ or $\pm C\sec^2\theta$; then $\to \frac{1}{2}\tan^2\theta$ or $\frac{1}{2}\sec^2\theta$ |
---
## Additional Notes for Part (c):
| Note |
|---|
| $\theta\sec^2\theta \to \theta\tan\theta + \ln(\sec\theta)$ **WITH NO INTERMEDIATE WORKING** is M0M0A0 |
| $\theta\sec^2\theta \to \theta\tan\theta - \ln(\cos\theta)$ **WITH NO INTERMEDIATE WORKING** is M0M0A0 |
| $\theta\sec^2\theta \to \theta\tan\theta - \ln(\sec\theta)$ **WITH NO INTERMEDIATE WORKING** is M1M1A1 |
| $\theta\sec^2\theta \to \theta\tan\theta + \ln(\cos\theta)$ **WITH NO INTERMEDIATE WORKING** is M1M1A1 |
| Alternative for $\int\tan\theta\sec^2\theta\,d\theta$: using $u=\tan\theta$, $\frac{du}{d\theta}=\sec^2\theta$, $\frac{dv}{d\theta}=\sec^2\theta$, $v=\tan\theta$ gives $\int\tan\theta\sec^2\theta\,d\theta = \tan^2\theta - \int\tan\theta\sec^2\theta\,d\theta \Rightarrow 2\int\tan\theta\sec^2\theta\,d\theta = \tan^2\theta \Rightarrow \int\tan\theta\sec^2\theta\,d\theta = \frac{1}{2}\tan^2\theta$ | M1, A1 |
| Or using $u=\sec\theta$: $\int\tan\theta\sec^2\theta\,d\theta = \sec^2\theta - \int\sec^2\theta\tan\theta\,d\theta \Rightarrow \int\tan\theta\sec^2\theta\,d\theta = \frac{1}{2}\sec^2\theta$ | M1, A1 |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd958ff3-ed4e-4bd7-aa4b-339da6d618a6-28_721_714_255_616}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 4 shows a sketch of part of the curve $C$ with parametric equations
$$x = 3 \theta \sin \theta , \quad y = \sec ^ { 3 } \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
The point $P ( k , 8 )$ lies on $C$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $k$.
The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the $y$-axis, the $x$-axis and the line with equation $x = k$.
\item Show that the area of $R$ can be expressed in the form
$$\lambda \int _ { \alpha } ^ { \beta } \left( \theta \sec ^ { 2 } \theta + \tan \theta \sec ^ { 2 } \theta \right) \mathrm { d } \theta$$
where $\lambda , \alpha$ and $\beta$ are constants to be determined.
\item Hence use integration to find the exact value of the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2017 Q8 [12]}}