| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Integrate using double angle |
| Difficulty | Standard +0.3 This is a standard double angle formula question requiring routine application of sin(2x) = 2sin(x)cos(x) and cos(2x) identities, followed by straightforward integration. Part (i) is algebraic manipulation using known identities, and part (ii) is direct integration of a constant and cosine term with given limits. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks |
|---|---|
| Use \(\sin 2x = 2\sin x \cos x\) to convert integrand to \(k \sin^2 2x\) | M1 |
| Use \(\cos 4x = 1 - 2 \sin^2 2x\) | M1 |
| State correct expression \(\frac{1}{2} - \frac{1}{2} \cos 4x\) or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(\cos^2 x = \frac{1 - \cos 2x}{2}\) and/or \(x = \frac{1 - \cos 2x}{2}\) to obtain an equation in \(\cos 2x\) only | M1 | |
| Use \(\cos^2 2x = \frac{1 + \cos 4x}{2}\) | M1 | |
| State correct expression \(\frac{1}{2} - \frac{1}{2} \cos 4x\) or equivalent | A1 | [3] |
| (ii) State correct integral \(\frac{3}{2}x - \frac{3}{8}\sin 4x\), or equivalent | B1 | |
| Attempt to substitute limits, using exact values | M1 | |
| Obtain given answer correctly | A1 | [3] |
(i) Either branch:
Use $\sin 2x = 2\sin x \cos x$ to convert integrand to $k \sin^2 2x$ | M1 |
Use $\cos 4x = 1 - 2 \sin^2 2x$ | M1 |
State correct expression $\frac{1}{2} - \frac{1}{2} \cos 4x$ or equivalent | A1 |
Or branch:
Use $\cos^2 x = \frac{1 - \cos 2x}{2}$ and/or $x = \frac{1 - \cos 2x}{2}$ to obtain an equation in $\cos 2x$ only | M1 |
Use $\cos^2 2x = \frac{1 + \cos 4x}{2}$ | M1 |
State correct expression $\frac{1}{2} - \frac{1}{2} \cos 4x$ or equivalent | A1 | [3]
(ii) State correct integral $\frac{3}{2}x - \frac{3}{8}\sin 4x$, or equivalent | B1 |
Attempt to substitute limits, using exact values | M1 |
Obtain given answer correctly | A1 | [3]3 (i) Show that $12 \sin ^ { 2 } x \cos ^ { 2 } x \equiv \frac { 3 } { 2 } ( 1 - \cos 4 x )$.\\
(ii) Hence show that
$$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } 12 \sin ^ { 2 } x \cos ^ { 2 } x d x = \frac { \pi } { 8 } + \frac { 3 \sqrt { } 3 } { 16 }$$
\hfill \mbox{\textit{CAIE P2 2013 Q3 [6]}}