Free practice / JC2 H2 Maths Prelim
Maclaurin Series: \(\cos(t^3)\); integrate; verify — NYJC 2025 H2 Math Prelim Paper 1
Attempt the question on paper, then open the worked solution to check your method.
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.
Show full worked solution▾
(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.