HCI 2025 H2 Math Prelim Paper 1 — Integration Techniques (worked solution)

Integration Techniques: IBP: \((2x-1)\cos x\); modulus — HCI 2025 H2 Math Prelim Paper 1

Hwa Chong Institution2025 PrelimPaper 1●●○ Standard7 marks

What this question tests

IBP: \((2x-1)\cos x\); modulus.

Question

(a) Find \(\displaystyle\int (2x-1)\cos x \,\mathrm{d}x\).

(b) Hence find the value of \(\displaystyle\int_{0}^{\pi/2} |2x-1|\cos x \,\mathrm{d}x\). Give your answer in the form \(A - 4\cos B\), where \(A\) and \(B\) are exact constants to be determined.

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(a)

Method 1

\[\begin{aligned} \int (2x-1)\cos x \,\mathrm{d}x &= (2x-1)\sin x - \int 2\sin x \,\mathrm{d}x \\ &= (2x-1)\sin x + 2\cos x + C \end{aligned}\]

Method 2

\[\begin{aligned} \int (2x-1)\cos x \,\mathrm{d}x &= \int 2x\cos x - \cos x \,\mathrm{d}x \\ &= 2x\sin x + \int 2\sin x \,\mathrm{d}x - \sin x + C \\ &= 2x\sin x - 2\cos x - \sin x + C \\ &= (2x-1)\sin x - 2\cos x + C \end{aligned}\]

(b)

Figure for this question

\[\begin{aligned} \int_{0}^{\pi/2} \lvert(2x-1)\rvert\cos x \,\mathrm{d}x &= -\int_{0}^{0.5}(2x-1)\cos x \,\mathrm{d}x + \int_{0.5}^{\pi/2}(2x-1)\cos x \,\mathrm{d}x \\ &= -\Big[(2x-1)\sin x + 2\cos x\Big]_{0}^{0.5} \\ &\phantom{{}={}} + \Big[(2x-1)\sin x + 2\cos x\Big]_{0.5}^{\pi/2} \\ &= -\Big[2\cos(0.5) - 2\Big] + \Big[(\pi-1)(1) + 2(0) - 2\cos(0.5)\Big] \\ &= \pi + 1 - 4\cos(0.5) \end{aligned}\] \(A = \pi + 1, \quad B = 0.5\)

Answer: (a) \((2x-1)\sin x + 2\cos x + C\) (b) \(\pi + 1 - 4\cos\tfrac{1}{2}\); \(A = \pi + 1\), \(B = \tfrac{1}{2}\)

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