2026 HCI P2 Q4
2026 HCI P2 Q4
Junior College 2
13 marks
- By expressing \(\dfrac{2x}{\sqrt{-x^2+2x}}=\dfrac{(ax-a)+a}{\sqrt{-x^2+2x}}\), where \(a\in\mathbb Z^+\), find \(\displaystyle\int\dfrac{2x}{\sqrt{-x^2+2x}}\,\mathrm dx\).[3]
- Show by integration that \(\displaystyle\int 2\sin^{-1}(x-1)\,\mathrm dx\) can be expressed as \[a(x-1)\sin^{-1}(x-1)+a\sqrt{2x-x^2}+C.\][3]
- Use the substitution \(u=\ln x\) to find \(\displaystyle\int\dfrac{\tan(\ln x)}{x}\,\mathrm dx\).[3]
- Hence find the exact area bounded by the graphs of \(y=2\sin^{-1}(x-1)\), \(y=\dfrac{\tan(\ln x)}{x}\), and the line \(x=1+\dfrac1{\sqrt2}\).[4]
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.
Create a free account to practise similar questions.
Answer:(a)(i) \(2\sin^{-1}(x-1)-2\sqrt{2x-x^2}+C\) (ii) \(2(x-1)\sin^{-1}(x-1)+2\sqrt{2x-x^2}+C\) (b) \(-\ln|\cos(\ln x)|+C\) (c) \(\frac{\pi}{2\sqrt2}+\sqrt2-2+\ln[\cos(\ln(1+1/\sqrt2))]\text{ units}^2\).
Need help? Join our JC Math tuition classes.
Learn more