Maclaurin Series: \(\cos(t^3)\); integrate; verify — NYJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
In the 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) \[ \cos(t^3) = 1 - \frac{(t^3)^2}{2!} + \frac{(t^3)^4}{4!} - \cdots = 1 - \frac{t^6}{2} + \frac{t^{12}}{24} + \cdots \]
\[\begin{aligned} \int_0^x t^2\cos(t^3)\;\mathrm{d}t &\approx \int_0^x t^2\!\left(1 - \frac{t^6}{2} + \frac{t^{12}}{24}\right)\!\mathrm{d}t\\ &= \int_0^x t^2 - \frac{t^8}{2} + \frac{t^{14}}{24}\;\mathrm{d}t\\ &= \left[\frac{t^3}{3} - \frac{t^9}{18} + \frac{t^{15}}{360}\right]_0^x\\ &= \frac{x^3}{3} - \frac{x^9}{18} + \frac{x^{15}}{360} \end{aligned}\](b) \[ \int_0^{0.1} t^2\cos(t^3)\;\mathrm{d}t \approx \frac{(0.1)^3}{3} - \frac{(0.1)^9}{18} + \frac{(0.1)^{15}}{360} \approx 0.00033\text{ (5 d.p.)} \]
(c) \[ \int_0^x t^2\cos(t^3)\;\mathrm{d}t = \tfrac{1}{3}\left[\sin(t^3)\right]_0^x = \tfrac{1}{3}\sin(x^3) \] \[ \int_0^{0.1} t^2\cos(t^3)\;\mathrm{d}t = \tfrac{1}{3}\sin\!\left(0.1^3\right) = 0.00033\text{ (5 d.p.)} \]
(d) The approximation was accurate to 5 d.p. The approximation was good since \(x = 0.1\) is close to zero.