| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Use trig identity before definite integration |
| Difficulty | Standard +0.3 Part (a) is a straightforward integration of cos(2x) using reverse chain rule. Part (b) requires knowing the identity tan²x = sec²x - 1, then integrating to get 3(tan x - x), which is standard technique. The exact value calculation with π limits is routine. This is slightly easier than average as it's a textbook application of a common identity with no problem-solving insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Obtain integral \(a \sin 2x\) with \(a = \pm \left(1, 2 \text{ or } \frac{1}{2}\right)\) | M1 | |
| Use limits and obtain \(\frac{1}{2}\) (AG) | A1 | [2] |
| (b) Use \(\tan^2 x = \sec^2 x - 1\) and attempt to integrate both terms | M1 | |
| Obtain \(3\tan x - 3x\) | A1 | |
| Attempt to substitute limits, using exact values | M1 | |
| Obtain answer \(2\sqrt{3} - \frac{\pi}{2}\) | A1 | [4] |
**(a)** Obtain integral $a \sin 2x$ with $a = \pm \left(1, 2 \text{ or } \frac{1}{2}\right)$ | M1 |
Use limits and obtain $\frac{1}{2}$ (AG) | A1 | [2]
**(b)** Use $\tan^2 x = \sec^2 x - 1$ and attempt to integrate both terms | M1 |
Obtain $3\tan x - 3x$ | A1 |
Attempt to substitute limits, using exact values | M1 |
Obtain answer $2\sqrt{3} - \frac{\pi}{2}$ | A1 | [4]4
\begin{enumerate}[label=(\alph*)]
\item Show that $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos 2 x \mathrm {~d} x = \frac { 1 } { 2 }$.
\item By using an appropriate trigonometrical identity, find the exact value of
$$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } 3 \tan ^ { 2 } x \mathrm {~d} x$$
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2010 Q4 [6]}}