| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Use trig identity before definite integration |
| Difficulty | Moderate -0.3 Part (a) is a standard bookwork proof of a double angle identity requiring minimal steps. Part (b) applies this identity to integrate cos²(3x), then evaluates definite integral—routine application of reverse chain rule with straightforward arithmetic. The question is slightly easier than average due to being heavily scaffolded and requiring only standard techniques. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05m Geometric proofs: of trig sum and double angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 2A \equiv \cos A\cos A - \sin A\sin A \equiv \cos^2 A - (1-\cos^2 A)\) | M1 | Writes \(\cos 2A \equiv \cos A\cos A - \sin A\sin A\) and attempts to use \(\pm\sin^2 A \pm \cos^2 A = \pm1\). Going directly to \(\cos 2A \equiv \cos^2 A - \sin^2 A\) in first step scores A0 |
| \(\Rightarrow \cos 2A \equiv 2\cos^2 A - 1\) | A1* | No errors seen, no circular reasoning. Correct notation throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(3 - 2\cos 6x)\,dx = 3x - \dfrac{\sin 6x}{3}\ (+c)\) | M1A1 | M1 for substituting part (a) result and integrating to form \(...x \pm ...\sin 6x\). \(\cos 2A \to k\sin 2A\) is M0 unless recovered |
| \(\left[3x - \dfrac{\sin 6x}{3}\right]_{\pi/12}^{\pi/8} = \dfrac{1}{8}\pi + \dfrac{2-\sqrt{2}}{6}\) | dM1A1 | dM1: substitutes both limits into expression of form \(...x\pm...\sin 6x\) and subtracts. A1: correct exact answer e.g. \(\dfrac{1}{8}\pi + \dfrac{1}{3} - \dfrac{\sqrt{2}}{6}\) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 2A \equiv \cos A\cos A - \sin A\sin A \equiv \cos^2 A - (1-\cos^2 A)$ | M1 | Writes $\cos 2A \equiv \cos A\cos A - \sin A\sin A$ and attempts to **use** $\pm\sin^2 A \pm \cos^2 A = \pm1$. Going directly to $\cos 2A \equiv \cos^2 A - \sin^2 A$ in first step scores A0 |
| $\Rightarrow \cos 2A \equiv 2\cos^2 A - 1$ | A1* | No errors seen, no circular reasoning. Correct notation throughout |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(3 - 2\cos 6x)\,dx = 3x - \dfrac{\sin 6x}{3}\ (+c)$ | M1A1 | M1 for substituting part (a) result and integrating to form $...x \pm ...\sin 6x$. $\cos 2A \to k\sin 2A$ is M0 unless recovered |
| $\left[3x - \dfrac{\sin 6x}{3}\right]_{\pi/12}^{\pi/8} = \dfrac{1}{8}\pi + \dfrac{2-\sqrt{2}}{6}$ | dM1A1 | dM1: substitutes both limits into expression of form $...x\pm...\sin 6x$ and subtracts. A1: correct exact answer e.g. $\dfrac{1}{8}\pi + \dfrac{1}{3} - \dfrac{\sqrt{2}}{6}$ |
---\begin{enumerate}
\item (a) Using the identity for $\cos ( A + B )$, prove that
\end{enumerate}
$$\cos 2 A \equiv 2 \cos ^ { 2 } A - 1$$
(b) Hence, using algebraic integration, find the exact value of
$$\int _ { \frac { \pi } { 12 } } ^ { \frac { \pi } { 8 } } \left( 5 - 4 \cos ^ { 2 } 3 x \right) d x$$
\hfill \mbox{\textit{Edexcel P3 2023 Q3 [6]}}