Maclaurin Series: binomial expansion, term-by-term integration and the \(\tan^{-1}\) series — JPJC 2025 H2 Math Prelim P1
What this question tests
Question
It is given that \(\displaystyle \mathrm{f}(x)=\frac{1}{4+9x^{2}}\).
(i) Find \(\displaystyle \int \mathrm{f}(x)\,\mathrm{d}x\).
(ii) 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.
(iii) Hence, find the Maclaurin series for \(\tan^{-1}\dfrac{3x}{2}\). Give the coefficients as exact fractions in their simplest form.
(iv) 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.
(v) Use your calculator to find \(\displaystyle \int_{0}^{0.5} \tan^{-1}\frac{3x}{2}\,\mathrm{d}x\), correct to 3 decimal places.
(vi) Comparing your answers to parts (iv) and (v), comment on the accuracy of your estimate in (iv) and how it can be improved.
Show full worked solution▾
(i) Integrate \(\mathrm{f}(x)\)
Factor out the \(9\) so the denominator matches the standard \(\dfrac{1}{a^2+x^2}\) inverse-tangent form.
\[ \begin{aligned} \int\frac{1}{4+9x^{2}}\,\mathrm{d}x &= \int\frac{1}{9\left(\dfrac{4}{9}+x^{2}\right)}\,\mathrm{d}x\\ &= \frac{1}{6}\tan^{-1}\frac{3x}{2}+K \end{aligned} \](ii) Binomial expansion of \(\mathrm{f}(x)\)
Pull out \(\tfrac14\) so the bracket is \((1+\text{small})^{-1}\), then apply the binomial series.
\[ \begin{aligned} \mathrm{f}(x) &= (4+9x^{2})^{-1} = \frac{1}{4}\left(1+\frac{9x^{2}}{4}\right)^{-1}\\ &= \frac{1}{4}\left[1+(-1)\left(\frac{9x^{2}}{4}\right)+\frac{-1(-2)}{2}\left(\frac{9x^{2}}{4}\right)^{2}+\cdots\right]\\ &= \frac{1}{4}\left(1-\frac{9x^{2}}{4}+\frac{81}{16}x^{4}+\cdots\right)\\ &\approx \frac{1}{4}-\frac{9}{16}x^{2}+\frac{81}{64}x^{4} \end{aligned} \](iii) Maclaurin series of \(\tan^{-1}\dfrac{3x}{2}\)
From part (i), \(\tan^{-1}\dfrac{3x}{2}\) is \(6\) times the integral of \(\mathrm{f}(x)\); integrate the expansion in (ii) term by term and fix the constant.
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) Estimate the integral using the series
Integrate the truncated series from \(0\) to \(0.5\) and evaluate.
\[ \begin{aligned} \int_{0}^{0.5}\tan^{-1}\frac{3x}{2}\,\mathrm{d}x &\approx \int_{0}^{0.5}\left(\frac{3}{2}x-\frac{9}{8}x^{3}+\frac{243}{160}x^{5}\right)\mathrm{d}x\\ &= 0.174 \quad\text{(to 3 d.p., using GC)} \end{aligned} \](v) Calculator value
Evaluate the exact integral directly on the graphing calculator.
\[ \int_{0}^{0.5}\tan^{-1}\frac{3x}{2}\,\mathrm{d}x = 0.173 \quad\text{(to 3 d.p.)} \](vi) Comment on accuracy
Compare the two values and identify what limits the precision.
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 \(\tan^{-1}\dfrac{3x}{2}\).