| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: show transformation then evaluate |
| Difficulty | Standard +0.8 This is a moderately challenging C4 integration question requiring trigonometric substitution to convert a surd into cos²θ, then applying double angle formulas. While the substitution is given, students must correctly handle the limits, derive k=4, and complete the integration using cos²θ = (1+cos2θ)/2. It's above average due to the multi-step nature and requirement to work with exact values, but follows a standard C4 pattern. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution |
| Answer | Marks |
|---|---|
| \(A = \int_0^{\frac{\pi}{6}}\sqrt{(3-x)(x+1)}dx\), \(x = 1 + 2\sin\theta\) | |
| \(\frac{dx}{d\theta} = 2\cos\theta\) | B1 |
| \(\frac{dx}{d\theta} = 2\cos\theta\) or \(2\cos\theta d\theta\). This mark can be implied by later working. Can be implied. | |
| \(\left[\int\sqrt{(3-x)(x+1)}dx\right.\) or \(\left.\int\sqrt{(3+2x-x^2)}dx\right]\) | |
| \(= \int\sqrt{(3-(1+2\sin\theta))(((1+2\sin\theta)+1)} 2\cos\theta \{d\theta\}\) | M1 |
| Substitutes for both \(x\) and \(dx_{,}\) where \(d\theta \neq \lambda d\theta\). Ignore \(d\theta\) | |
| \(= \int\sqrt{(2-2\sin\theta)(2+2\sin\theta)} 2\cos\theta \{d\theta\}\) | |
| \(= \int\sqrt{4-4\sin^2\theta} 2\cos\theta \{d\theta\}\) | |
| \(= \int\sqrt{4\cos^2\theta} 2\cos\theta \{d\theta\}\) or \(\int\sqrt{4\cos^2\theta} 2\cos\theta \{d\theta\}\) | M1 |
| Applies \(\cos^2\theta = 1 - \sin^2\theta\) or see notes. | |
| \(= 4\int\cos^2\theta d\theta\), \(\{k=4\}\) | A1 |
| \(4\cos^2\theta d\theta\) or \(\int 4\cos^2\theta d\theta\) | |
| Note: \(d\theta\) is required here. | |
| \(0 = 1 + 2\sin\theta\) or \(-1 = 2\sin\theta\) or \(\sin\theta = -\frac{1}{2} \Rightarrow \theta = -\frac{\pi}{6}\) | B1 |
| and \(3 = 1 + 2\sin\theta\) or \(2 = 2\sin\theta\) or \(\sin\theta = 1 \Rightarrow \theta = \frac{\pi}{2}\) | |
| [5] |
| Answer | Marks |
|---|---|
| \(\left\{k\int\cos^2\theta\{d\theta\}\right\} = \{k\}\left\{\int\left(\frac{1+\cos 2\theta}{2}\right)\{d\theta\}\right\}\) | M1 |
| Applies \(\cos^2\theta = 1 - \sin^2\theta\) to their integral | |
| \(= \{k\}\left\{\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right\}\) or \(k(\alpha\theta \pm \beta\sin 2\theta)\), \(\alpha \neq 0, \beta \neq 0\) | M1 (A1 on ePEN) |
| Integrates to give an expression of the form \(\pm \alpha\theta \pm \beta\sin 2\theta\), \(\alpha \neq 0, \beta \neq 0\) (can be simplified or un-simplified). | |
| So \(4\int\cos^2\theta d\theta = \left[2\theta + \sin 2\theta\right]_{-\frac{\pi}{6}}^{\frac{\pi}{2}}\) | |
| \(= \left(2\left(\frac{\pi}{2}\right)+\sin\left(\frac{2\pi}{2}\right)\right)-\left(2\left(-\frac{\pi}{6}\right)+\sin\left(-\frac{2\pi}{6}\right)\right)\) | |
| \(= (\pi) - \left(-\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\) | A1 |
| \(= \frac{4\pi}{3} + \frac{\sqrt{3}}{2}\) or \(\frac{4\pi}{3} + \frac{\sqrt{3}}{2}\) or \(\frac{1}{6}(8\pi + 3\sqrt{3})\) | cso A1 |
| [3] |
## Part (a)
$A = \int_0^{\frac{\pi}{6}}\sqrt{(3-x)(x+1)}dx$, $x = 1 + 2\sin\theta$ | |
$\frac{dx}{d\theta} = 2\cos\theta$ | B1 |
$\frac{dx}{d\theta} = 2\cos\theta$ or $2\cos\theta d\theta$. This mark can be implied by later working. Can be implied. | |
$\left[\int\sqrt{(3-x)(x+1)}dx\right.$ or $\left.\int\sqrt{(3+2x-x^2)}dx\right]$ | |
$= \int\sqrt{(3-(1+2\sin\theta))(((1+2\sin\theta)+1)} 2\cos\theta \{d\theta\}$ | M1 |
Substitutes for both $x$ and $dx_{,}$ where $d\theta \neq \lambda d\theta$. Ignore $d\theta$ | |
$= \int\sqrt{(2-2\sin\theta)(2+2\sin\theta)} 2\cos\theta \{d\theta\}$ | |
$= \int\sqrt{4-4\sin^2\theta} 2\cos\theta \{d\theta\}$ | |
$= \int\sqrt{4\cos^2\theta} 2\cos\theta \{d\theta\}$ or $\int\sqrt{4\cos^2\theta} 2\cos\theta \{d\theta\}$ | M1 |
Applies $\cos^2\theta = 1 - \sin^2\theta$ or see notes. | |
$= 4\int\cos^2\theta d\theta$, $\{k=4\}$ | A1 |
$4\cos^2\theta d\theta$ or $\int 4\cos^2\theta d\theta$ | |
Note: $d\theta$ is required here. | |
$0 = 1 + 2\sin\theta$ or $-1 = 2\sin\theta$ or $\sin\theta = -\frac{1}{2} \Rightarrow \theta = -\frac{\pi}{6}$ | B1 |
and $3 = 1 + 2\sin\theta$ or $2 = 2\sin\theta$ or $\sin\theta = 1 \Rightarrow \theta = \frac{\pi}{2}$ | |
| [5] |
## Part (b)
$\left\{k\int\cos^2\theta\{d\theta\}\right\} = \{k\}\left\{\int\left(\frac{1+\cos 2\theta}{2}\right)\{d\theta\}\right\}$ | M1 |
Applies $\cos^2\theta = 1 - \sin^2\theta$ to their integral | |
$= \{k\}\left\{\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right\}$ or $k(\alpha\theta \pm \beta\sin 2\theta)$, $\alpha \neq 0, \beta \neq 0$ | M1 (A1 on ePEN) |
Integrates to give an expression of the form $\pm \alpha\theta \pm \beta\sin 2\theta$, $\alpha \neq 0, \beta \neq 0$ (can be simplified or un-simplified). | |
So $4\int\cos^2\theta d\theta = \left[2\theta + \sin 2\theta\right]_{-\frac{\pi}{6}}^{\frac{\pi}{2}}$ | |
$= \left(2\left(\frac{\pi}{2}\right)+\sin\left(\frac{2\pi}{2}\right)\right)-\left(2\left(-\frac{\pi}{6}\right)+\sin\left(-\frac{2\pi}{6}\right)\right)$ | |
$= (\pi) - \left(-\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)$ | A1 |
$= \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$ or $\frac{4\pi}{3} + \frac{\sqrt{3}}{2}$ or $\frac{1}{6}(8\pi + 3\sqrt{3})$ | cso A1 |
| [3] |
---\includegraphics{figure_2}
Figure 2 shows a sketch of the curve with equation $y = \sqrt{(3-x)(x+1)}$, $0 \leqslant x \leqslant 3$
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis, and the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $x = 1 + 2\sin\theta$ to show that
$$\int_0^3 \sqrt{(3-x)(x+1)} dx = k \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2\theta d\theta$$
where $k$ is a constant to be determined.
[5]
\item Hence find, by integration, the exact area of $R$.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q6 [8]}}