CAIE FP1 2009 November — Question 7 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyChallenging +1.2 This is a standard Further Maths question requiring systematic application of de Moivre's theorem to express sin^6θ in terms of multiple angles, followed by a straightforward integration with substitution. While it involves multiple steps and algebraic manipulation, the technique is well-practiced in FP1 and follows a predictable pattern without requiring novel insight.
Spec1.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits4.02q De Moivre's theorem: multiple angle formulae

7 Use de Moivre's theorem to express \(\sin ^ { 6 } \theta\) in the form $$a + b \cos 2 \theta + c \cos 4 \theta + d \cos 6 \theta$$ where \(a , b , c , d\) are constants to be found. Hence evaluate $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } 2 x d x$$ leaving your answer in terms of \(\pi\).

AnswerMarks
\(64\sin^6\theta = -(z - 1/z)^6 (z = e^{i\theta})\)M1
\(= -(z^6 + 1/z^6) + 6(z^4 + 1/z^4) - 15(z^2 + 1/z^2) + 20\)M1A1
\(\sin^6\theta = 5/16 - (15/32)\cos 2\theta + (3/16)\cos 4\theta - (1/32)\cos 6\theta\)M1A1
\(\sin^6 2x = 5/16 - (15/32)\cos 4x + (3/16)\cos 8x - (1/32)\cos 12x\)M1
Any one of \(\int_0^{\pi/4} \cos kx \, dx = 0\) for \(k = 1, 2, 3\)B1
Further B1 if all 3 are written down or impliedB1
\(\int_0^{\pi/4} \sin^6 2x \, dx = 5\pi/64\) or \(\pi a/4\) (CWO)A1
OR for last 4 marks
\(I = (1/2)\int_0^{\pi/2} \sin^6 u \, du = (1/64)\int_0^{\pi/2} (10 - 15\cos 2u + 6\cos 4u - \cos 6u) du\)M1
\(= (1/64)[10u - 15\sin 2u/2 + 3\sin 4u/2 - \sin 6u/6]_0^{\pi/2}\)M1A1
\(= 5\pi/64\) (CWO)A1
$64\sin^6\theta = -(z - 1/z)^6 (z = e^{i\theta})$ | M1 |
$= -(z^6 + 1/z^6) + 6(z^4 + 1/z^4) - 15(z^2 + 1/z^2) + 20$ | M1A1 |
$\sin^6\theta = 5/16 - (15/32)\cos 2\theta + (3/16)\cos 4\theta - (1/32)\cos 6\theta$ | M1A1 |
$\sin^6 2x = 5/16 - (15/32)\cos 4x + (3/16)\cos 8x - (1/32)\cos 12x$ | M1 |
Any one of $\int_0^{\pi/4} \cos kx \, dx = 0$ for $k = 1, 2, 3$ | B1 |
Further B1 if all 3 are written down or implied | B1 |
$\int_0^{\pi/4} \sin^6 2x \, dx = 5\pi/64$ or $\pi a/4$ (CWO) | A1 |

**OR** for last 4 marks | |
$I = (1/2)\int_0^{\pi/2} \sin^6 u \, du = (1/64)\int_0^{\pi/2} (10 - 15\cos 2u + 6\cos 4u - \cos 6u) du$ | M1 |
$= (1/64)[10u - 15\sin 2u/2 + 3\sin 4u/2 - \sin 6u/6]_0^{\pi/2}$ | M1A1 |
$= 5\pi/64$ (CWO) | A1 |
7 Use de Moivre's theorem to express $\sin ^ { 6 } \theta$ in the form

$$a + b \cos 2 \theta + c \cos 4 \theta + d \cos 6 \theta$$

where $a , b , c , d$ are constants to be found.

Hence evaluate

$$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } 2 x d x$$

leaving your answer in terms of $\pi$.

\hfill \mbox{\textit{CAIE FP1 2009 Q7 [9]}}