| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Integration using reciprocal identities |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard reciprocal trig identities and basic integration. Part (i) is routine algebraic manipulation using cot θ = cos θ/sin θ and tan θ = sin θ/cos θ, then applying the double angle formula. Part (ii) follows directly from (i) with a simple substitution and logarithmic integration. While it requires multiple techniques, each step is standard bookwork with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05p Proof involving trig: functions and identities1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use Pythagoras | M1 | |
| Use the \(\sin 2A\) formula | M1 | |
| Obtain the given result | A1 | [3] |
| (ii) Integrate and obtain a \(k\ln\sin\theta\) or \(m\ln\cos\theta\) term, or obtain integral of the form \(p\ln\tan\theta\) | M1* | |
| Obtain indefinite integral \(\frac{1}{2}\ln\sin\theta-\frac{1}{2}\ln\cos\theta\), or equivalent, or \(\frac{1}{2}\ln\tan\theta\) | A1 | |
| Substitute limits correctly | M1(dep)* | |
| Obtain the given answer correctly having shown appropriate working | A1 | [4] |
**(i)** Use Pythagoras | M1 |
Use the $\sin 2A$ formula | M1 |
Obtain the given result | A1 | [3]
**(ii)** Integrate and obtain a $k\ln\sin\theta$ or $m\ln\cos\theta$ term, or obtain integral of the form $p\ln\tan\theta$ | M1* |
Obtain indefinite integral $\frac{1}{2}\ln\sin\theta-\frac{1}{2}\ln\cos\theta$, or equivalent, or $\frac{1}{2}\ln\tan\theta$ | A1 |
Substitute limits correctly | M1(dep)* |
Obtain the given answer correctly having shown appropriate working | A1 | [4]5 (i) Prove that $\cot \theta + \tan \theta \equiv 2 \operatorname { cosec } 2 \theta$.\\
(ii) Hence show that $\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \operatorname { cosec } 2 \theta \mathrm {~d} \theta = \frac { 1 } { 2 } \ln 3$.
\hfill \mbox{\textit{CAIE P3 2013 Q5 [7]}}