| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Difficulty | Moderate -0.3 This is a straightforward two-part integration question requiring standard techniques: part (a) uses the identity tan²x = sec²x - 1 (routine manipulation) and sin 2x integrates directly; part (b) is a simple exponential integration with substitution. Both parts are textbook exercises with no problem-solving required, making this slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Use \(\tan^2 x = \sec^2 x - 1\). Obtain integral of form \(p \tan x + qx + r \cos 2x\) | B1 M1 | |
| Obtain \(\tan x - x - \frac{1}{2} \cos 2x + c\) | A1 | [3] |
| (b) Obtain integral of form \(ke^{-2x}\) | M1* | |
| Obtain \(-\frac{3}{2}e^{-2x}\) | A1 | |
| Apply both limits the correct way round. Obtain \(-\frac{3}{2}e^{-1} + \frac{3}{2}e\) or exact equivalent | M1 dep A1 | [4] |
(a) Use $\tan^2 x = \sec^2 x - 1$. Obtain integral of form $p \tan x + qx + r \cos 2x$ | B1 M1 |
Obtain $\tan x - x - \frac{1}{2} \cos 2x + c$ | A1 | [3]
(b) Obtain integral of form $ke^{-2x}$ | M1* |
Obtain $-\frac{3}{2}e^{-2x}$ | A1 |
Apply both limits the correct way round. Obtain $-\frac{3}{2}e^{-1} + \frac{3}{2}e$ or exact equivalent | M1 dep A1 | [4]5
\begin{enumerate}[label=(\alph*)]
\item Find $\int \left( \tan ^ { 2 } x + \sin 2 x \right) \mathrm { d } x$.
\item Find the exact value of $\int _ { 0 } ^ { 1 } 3 \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2015 Q5 [7]}}