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Integration from AEA

2)
a) Show that sin3x3sinx4sin3x\sin 3x \equiv 3\sin x - 4\sin^3x (3 marks)
I have done this.

Hence find
b) cosx(6sinx2sin3x)23dx\int \cos x(6\sin x - 2\sin 3x)^{\frac{2}{3}} dx (3 marks)
I have done this.

c) (3sin2x2sin3xcosx)13dx\int (3\sin 2x - 2\sin 3x \cos x)^{\frac{1}{3}} dx (4 marks)

I have simplified the integrand to (8sin3xcosx)13dx\int (8\sin^3x\cos x)^{\frac{1}{3}} dx but I am unsure of how to proceed.
Original post by Quivai
2)
a) Show that sin3x3sinx4sin3x\sin 3x \equiv 3\sin x - 4\sin^3x (3 marks)
I have done this.

Hence find
b) cosx(6sinx2sin3x)23dx\int \cos x(6\sin x - 2\sin 3x)^{\frac{2}{3}} dx (3 marks)
I have done this.

c) (3sin2x2sin3xcosx)13dx\int (3\sin 2x - 2\sin 3x \cos x)^{\frac{1}{3}} dx (4 marks)

I have simplified the integrand to (8sin3xcosx)13dx\int (8\sin^3x\cos x)^{\frac{1}{3}} dx but I am unsure of how to proceed.

(8sin3xcosx)132(cosx)13sinx(8\sin ^3x\cos x)^{\frac{1}{3}} \equiv 2(\cos x)^{\frac{1}{3}}\sin x. Can you see an obvious substitution from this form of the integrand?
Hint: ddx[cosx]=sinx\dfrac{d}{dx}[\cos x] = -\sin x
Reply 2
How do you do the first part? I have done it via De Moivre's theorem (FP2-Edexcel) but am unsure how to do it in a Core Maths way
Reply 3
Original post by Gome44
How do you do the first part? I have done it via De Moivre's theorem (FP2-Edexcel) but am unsure how to do it in a Core Maths way


You mean part 2a - working out sin(3x)?

That's just a straightforward application of the addition formula for sin(2x + x), with a conversion of cos2x into sin^2x thrown in as well :smile:
Reply 4
Ok, cool.

I always over-complicate things

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