Given that ${y = \ln{(\sec{x})}}$, show that $\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 2 \left( \frac{

Question

Given that ${y = \ln{(\sec{x})}}$, show that $\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 2 \left( \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} ight) \left( \frac{\mathrm{d}y}{\mathrm{d}x} ight)$. [3] Hence or otherwise, obtain the Maclaurin expansion of ${y}$ in terms of ${x}$ up to and including the term in ${x^4}$. [3] By putting ${x = \frac{\pi}{4}}$, find an approximation for ${\ln{2}}$ in terms of ${\pi}$. [2] Using your answer to part (b), find an approximation to $\displaystyle\int
olimits_{0}^{\frac{\pi }{10}}{\ln }(\sec x)\mathrm{d}x$. Give your answer correct to 4 significant figures. [1]

Accepted Answer

(a) shown; (b) y = x^2/2 + x^4/12 + ...; (c) ln 2 = pi^2/16 + pi^4/1536; (d) 0.005219

N2023 P2 Q2

N2023 P2 Q2

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