Maclaurin Series: \(1/(4+9x^2)\); \(\arctan\) series — JPJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
It is given that \(\displaystyle \mathrm{f}(x)=\frac{1}{4+9x^{2}}\).
- Find \(\displaystyle \int \mathrm{f}(x)\,\mathrm{d}x\).
- Find the binomial expansion for \(\mathrm{f}(x)\), up to and including the term in \(x^{4}\). Give the coefficients as exact fractions in their simplest form.
- Hence, find the Maclaurin series for \(\tan^{-1}\dfrac{3x}{2}\). Give the coefficients as exact fractions in their simplest form.
- Use your series from part (iii) to estimate \(\displaystyle \int_{0}^{0.5} \tan^{-1}\frac{3x}{2}\,\mathrm{d}x\), correct to 3 decimal places.
- Use your calculator to find \(\displaystyle \int_{0}^{0.5} \tan^{-1}\frac{3x}{2}\,\mathrm{d}x\), correct to 3 decimal places.
- Comparing your answers to parts (iv) and (v), comment on the accuracy of your estimate in (iv) and how it can be improved.
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(i) [2 marks] \[\begin{aligned} \int\frac{1}{4+9x^{2}}\,\mathrm{d}x &= \int\frac{1}{9\!\left(\dfrac{4}{9}+x^{2}\right)}\,\mathrm{d}x\\ &= \frac{1}{6}\tan^{-1}\!\frac{3x}{2}+K \end{aligned}\]
(ii) [3 marks] \[\begin{aligned} \mathrm{f}(x) &= (4+9x^{2})^{-1} = \frac{1}{4}\!\left(1+\frac{9x^{2}}{4}\right)^{\!-1}\\ &= \frac{1}{4}\!\left[1+(-1)\!\left(\frac{9x^{2}}{4}\right)+\frac{-1(-2)}{2}\!\left(\frac{9x^{2}}{4}\right)^{\!2}+\cdots\right]\\ &= \frac{1}{4}\!\left(1-\frac{9x^{2}}{4}+\frac{81}{16}x^{4}+\cdots\right)\\ &\approx \frac{1}{4}-\frac{9}{16}x^{2}+\frac{81}{64}x^{4} \end{aligned}\]
(iii) [3 marks]
From (i): \(\tan^{-1}\!\dfrac{3x}{2} = 6\displaystyle\int\frac{1}{4+9x^{2}}\,\mathrm{d}x + C\), where \(C = -6K\).
From (ii): \[\begin{aligned} \tan^{-1}\!\frac{3x}{2} &= 6\!\int\!\left(\frac{1}{4}-\frac{9}{16}x^{2}+\frac{81}{64}x^{4}\right)\mathrm{d}x + C\\ &= 6\!\left(\frac{1}{4}x-\frac{3}{16}x^{3}+\frac{81}{320}x^{5}\right)+D\\ &= \frac{3}{2}x-\frac{9}{8}x^{3}+\frac{243}{160}x^{5}+D \end{aligned}\]
When \(x = 0\), \(\tan^{-1}0 = 0 \;\Rightarrow\; D = 0\). \[ \therefore\; \tan^{-1}\!\frac{3x}{2} = \frac{3}{2}x-\frac{9}{8}x^{3}+\frac{243}{160}x^{5}+\cdots \]
(iv) [1 mark] \[\begin{aligned} \int_{0}^{0.5}\tan^{-1}\!\frac{3x}{2}\,\mathrm{d}x &\approx \int_{0}^{0.5}\!\left(\frac{3}{2}x-\frac{9}{8}x^{3}+\frac{243}{160}x^{5}\right)\mathrm{d}x\\ &= 0.174 \quad\text{(to 3 d.p.)}\quad(\text{using GC}) \end{aligned}\]
(v) [1 mark]
From GC: \(\displaystyle\int_{0}^{0.5}\tan^{-1}\!\frac{3x}{2}\,\mathrm{d}x = 0.173\) (to 3 d.p.).
(vi) [1 mark]
The estimate in (iv) is accurate to 2 decimal places but not to 3 decimal places. To improve the estimate, include higher-order terms in the Maclaurin series expansion of \(\tan^{-1}\!\dfrac{3x}{2}\).