CAIE P2 2008 June — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2008
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with trigonometric functions
DifficultyModerate -0.8 This is a straightforward definite integral requiring only standard integration formulas for cos(2x) and sin(x), followed by substitution of limits. The reverse chain rule for cos(2x) is routine, and evaluating at the given limits involves standard exact values (π/6). This is simpler than an average A-level question as it requires no problem-solving—just direct application of memorized integration rules.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

3 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( \cos 2 x + \sin x ) \mathrm { d } x\).

AnswerMarks
Obtain integral \(\frac{1}{2}\sin 2x - \cos x\)B1 + B1
Substitute limits correctly in an integral of the form \(a\sin 2x + b\cos x\)M1
Use correct exact values, e.g. of \(\cos\left(\frac{\pi}{6}\right)\)M1
Obtain answer \(1 - \frac{1}{4}\sqrt{3}\), or equivalentA1
[5]
Obtain integral $\frac{1}{2}\sin 2x - \cos x$ | B1 + B1 |
Substitute limits correctly in an integral of the form $a\sin 2x + b\cos x$ | M1 |
Use correct exact values, e.g. of $\cos\left(\frac{\pi}{6}\right)$ | M1 |
Obtain answer $1 - \frac{1}{4}\sqrt{3}$, or equivalent | A1 |
| [5] |
3 Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( \cos 2 x + \sin x ) \mathrm { d } x$.

\hfill \mbox{\textit{CAIE P2 2008 Q3 [5]}}