It is given that $y=\frac{1}{3+\sin 2x}$. Show that \(\frac{{{\mathrm{d}}^{2}}y}{\mathrm

Question

It is given that $y=\frac{1}{3+\sin 2x}$. Show that $\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-4{{y}^{2}}\sin 2x+4y\frac{\mathrm{d}y}{\mathrm{d}x}\cos 2x=0$. Hence find the Maclaurin series for $y$, up to and including the term in ${{x}^{2}}$.[4] Given further that $x$ is small so that ${{x}^{3}}$ and higher powers of $x$ can be neglected, use appropriate expansions from the List of Formulae (MF26) to verify the correctness of the series of $y$ in part (a). [3] Use the series expansion found in part (a) and integration to find an approximate value for $\int_0^{1.5} \frac{1}{3+\sin 2x}\mathrm{d}x.$. [2]

Accepted Answer

(a) ${\frac{1}{3} - \frac{2}{9}x + \frac{4}{27}x^2}$ (c) ${\frac{5}{12}}$ or ${0.417}$

2023 ASRJC P1 Q8

2023 ASRJC P1 Q8

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