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2 marksIntegration II

CAPE Pure Mathematics Unit 2 · May/June 2012 · Paper 2 · Question 2(b)(i)

Given that \sin A \cos B - \cos A \sin B = \sin(A - B), show that \cos 3x \sin x = \sin 3x \cos x - \sin 2x.

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Other parts of this question

  1. 2(a)(i)Express \frac{x^2 - 3x}{(x - 1)(x^2 + 1)} in partial fractions.[7 marks]
  2. 2(a)(ii)Hence, find \int \frac{x^2 - 3x}{x^3 - x^2 + x - 1} \,\mathrm{d}x.[5 marks]
  3. 2(b)(ii)Prove that (m + 3) I_m = m J_{m-1} - \cos^m x \cos 3x.[7 marks]
  4. 2(b)(iii)Hence, by setting m = 1, prove that 4 \int_0^{\frac{\pi}{4}} \cos x \sin 3x \,\mathrm{d}x = \int_0^{\frac{\pi}{4}} \sin 2x \,\mathrm{d}x + \frac{3}{2}.[2 marks]
  5. 2(b)(iv)Evaluate \int_0^{\frac{\pi}{4}} \sin 2x \,\mathrm{d}x.[2 marks]

More practice: the rest of this paper · more Integration II questions · all CAPE Pure Mathematics Unit 2 past papers