Edexcel AEA 2012 June — Question 2 10 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
TopicIntegration by Substitution
TypeSubstitution u = sin x or u = cos x (area/integral)
DifficultyChallenging +1.8 This AEA question requires proving a triple-angle identity, then applying it cleverly to simplify two non-standard integrals. Part (b) requires recognizing that the integrand simplifies to a power of sin x using the identity, making substitution straightforward. Part (c) is more sophisticated, requiring manipulation of the expression using both the proven identity and double-angle formulas before integration. While systematic, it demands strong algebraic manipulation and pattern recognition beyond typical A-level integration questions.
Spec1.05l Double angle formulae: and compound angle formulae1.08h Integration by substitution

2.(a)Show that $$\sin 3 x = 3 \sin x - 4 \sin ^ { 3 } x$$ Hence find
(b) \(\int \cos x ( 6 \sin x - 2 \sin 3 x ) ^ { \frac { 2 } { 3 } } \mathrm {~d} x\) (c) \(\int ( 3 \sin 2 x - 2 \sin 3 x \cos x ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\)

2.(a)Show that

$$\sin 3 x = 3 \sin x - 4 \sin ^ { 3 } x$$

Hence find\\
(b) $\int \cos x ( 6 \sin x - 2 \sin 3 x ) ^ { \frac { 2 } { 3 } } \mathrm {~d} x$\\
(c) $\int ( 3 \sin 2 x - 2 \sin 3 x \cos x ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x$

\hfill \mbox{\textit{Edexcel AEA 2012 Q2 [10]}}