Maclaurin Series: expansion of \(\cos(t^3)\), term-by-term integration and verification — NYJC 2025 H2 Math Prelim P1
What this question tests
Question
In this question you may use expansions from the List of Formulae (MF27).
(a) Find the Maclaurin expansion of \(\cos(t^3)\) in ascending powers of \(t\), up to and including the term in \(t^{12}\). Hence, find the Maclaurin series of \(\displaystyle\int_0^x t^2\cos(t^3)\,\mathrm{d}t\) up to and including the term in \(x^{15}\), given that \(x\) is small.
(b) Use your expansion from part (a) to find an approximate value for \(\displaystyle\int_0^{0.1} t^2\cos(t^3)\,\mathrm{d}t\), correct to 5 decimal places.
(c) Find \(\displaystyle\int_0^x t^2\cos(t^3)\,\mathrm{d}t\) in terms of \(x\). Hence evaluate \(\displaystyle\int_0^{0.1} t^2\cos(t^3)\,\mathrm{d}t\), correct to 5 decimal places.
(d) Comparing your answers to parts (b) and (c), and with reference to the value of \(x\), comment on the accuracy of your approximations.
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(a) Expansion and term-by-term integration
Substitute \(t^3\) into the standard \(\cos u\) series, multiply by \(t^2\), then integrate each power.
\[ \cos(t^3) = 1 - \frac{(t^3)^2}{2!} + \frac{(t^3)^4}{4!} - \cdots = 1 - \frac{t^6}{2} + \frac{t^{12}}{24} + \cdots \] \[ \begin{aligned} \int_0^x t^2\cos(t^3)\;\mathrm{d}t &\approx \int_0^x t^2\!\left(1 - \frac{t^6}{2} + \frac{t^{12}}{24}\right)\!\mathrm{d}t\\ &= \int_0^x t^2 - \frac{t^8}{2} + \frac{t^{14}}{24}\;\mathrm{d}t\\ &= \left[\frac{t^3}{3} - \frac{t^9}{18} + \frac{t^{15}}{360}\right]_0^x\\ &= \frac{x^3}{3} - \frac{x^9}{18} + \frac{x^{15}}{360} \end{aligned} \](b) Approximation using the series
Substitute \(x = 0.1\) into the polynomial from part (a).
\[ \int_0^{0.1} t^2\cos(t^3)\;\mathrm{d}t \approx \frac{(0.1)^3}{3} - \frac{(0.1)^9}{18} + \frac{(0.1)^{15}}{360} \approx 0.00033 \text{ (5 d.p.)} \](c) Exact integral and exact evaluation
Recognise that \(t^2\cos(t^3)\) is \(\tfrac{1}{3}\) the derivative of \(\sin(t^3)\), so the integral is exact.
\[ \int_0^x t^2\cos(t^3)\;\mathrm{d}t = \tfrac{1}{3}\left[\sin(t^3)\right]_0^x = \tfrac{1}{3}\sin(x^3) \] \[ \int_0^{0.1} t^2\cos(t^3)\;\mathrm{d}t = \tfrac{1}{3}\sin\!\left(0.1^3\right) = 0.00033 \text{ (5 d.p.)} \](d) Comment on accuracy
Compare the two values in light of how small \(x\) is.
The approximation in (b) agrees with the exact value in (c) to 5 decimal places. The approximation is good because \(x = 0.1\) is close to zero, so the neglected higher-order terms are negligibly small.