Edexcel AEA 2012 June — Question 2

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2012
SessionJune
TopicIntegration 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) \(\int ( 3 \sin 2 x - 2 \sin 3 x \cos x ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x\)