已知 $y = \dfrac{1}{3 + \sin 2x}$。

1. Rewrite the equation to avoid fractions and find the first derivative implicitly with respect to $x$.

$y = \frac{1}{3 + \sin 2x}$
$y(3 + \sin 2x) = 1$
Differentiating with respect to $x$ using the product rule:
$\frac{\mathrm{d}y}{\mathrm{d}x}(3 + \sin 2x) + y(2\cos 2x) = 0$

2. Differentiate a second time to show the required differential equation.

Differentiating implicitly again with respect to $x$:
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}(3 + \sin 2x) + \frac{\mathrm{d}y}{\mathrm{d}x}(2\cos 2x) + \frac{\mathrm{d}y}{\mathrm{d}x}(2\cos 2x) + y(-4\sin 2x) = 0$
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}(3 + \sin 2x) + 4\frac{\mathrm{d}y}{\mathrm{d}x}\cos 2x - 4y\sin 2x = 0$
Since $y = \frac{1}{3 + \sin 2x}$, we can substitute $(3 + \sin 2x) = \frac{1}{y}$:
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\left(\frac{1}{y}\right) + 4\frac{\mathrm{d}y}{\mathrm{d}x}\cos 2x - 4y\sin 2x = 0$
Multiplying the entire equation by $y$ gives the required result:
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 4y^2 \sin 2x + 4y\frac{\mathrm{d}y}{\mathrm{d}x} \cos 2x = 0 \quad \text{(shown)}$

3. Evaluate $y$, $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x = 0$.

When $x = 0$:
$y = \frac{1}{3 + \sin 0} = \frac{1}{3}$
Substituting into the first derivative equation:
$\frac{\mathrm{d}y}{\mathrm{d}x}(3 + 0) + \left(\frac{1}{3}\right)(2\cos 0) = 0 \implies 3\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{3} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2}{9}$
Substituting into the second derivative equation:
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 4\left(\frac{1}{3}\right)^2(0) + 4\left(\frac{1}{3}\right)\left(-\frac{2}{9}\right)(1) = 0$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 0 - \frac{8}{27} = 0 \implies \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{8}{27}$

4. Construct the Maclaurin series for $y$ up to the $x^2$ term.

Using the Maclaurin series formula $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$ :
$y = \frac{1}{3} + \left(-\frac{2}{9}\right)x + \frac{\frac{8}{27}}{2}x^2 + \dots$
$y = \frac{1}{3} - \frac{2}{9}x + \frac{4}{27}x^2$

1. Substitute the standard series expansion for $\sin 2x$.

From the List of Formulae (MF26), the standard series expansion for $\sin u$ is $u - \frac{u^3}{3!} + \dots$
Since $x^3$ and higher powers of $x$ can be neglected, $\sin 2x \approx 2x$.
Therefore,
$y \approx \frac{1}{3 + 2x}$

2. Rewrite the expression into a form suitable for the binomial series expansion.

$y \approx (3 + 2x)^{-1}$
Factoring out the constant to get the form $(1 + u)^{-1}$:
$y \approx \left[ 3 \left( 1 + \frac{2x}{3} \right) \right]^{-1}$
$y \approx \frac{1}{3} \left( 1 + \frac{2x}{3} \right)^{-1}$

3. Apply the binomial series expansion $(1+u)^n$ up to the term in $u^2$.

From MF26, the binomial expansion for any rational index is $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$
Applying this with $n = -1$ and $u = \frac{2x}{3}$:
$y \approx \frac{1}{3} \left[ 1 + (-1)\left(\frac{2x}{3}\right) + \frac{(-1)(-2)}{2!} \left(\frac{2x}{3}\right)^2 \right]$

4. Simplify the expansion to verify the series from part (a).

$y \approx \frac{1}{3} \left( 1 - \frac{2x}{3} + \frac{4x^2}{9} \right)$
$y \approx \frac{1}{3} - \frac{2}{9}x + \frac{4}{27}x^2$
This matches and verifies the correctness of the Maclaurin series obtained in part (a).

1. Substitute the Maclaurin series for $y$ from part (a) into the definite integral.

From part (a), we have the expansion $\dfrac{1}{3 + \sin 2x} \approx \dfrac{1}{3} - \dfrac{2}{9}x + \dfrac{4}{27}x^2$.

$\int_{0}^{1.5} \dfrac{1}{3 + \sin 2x}\;\mathrm{d}x \approx \int_{0}^{1.5} \left( \dfrac{1}{3} - \dfrac{2}{9}x + \dfrac{4}{27}x^2 \right)\;\mathrm{d}x$

2. Integrate the polynomial and evaluate at the limits $0$ and $1.5$.

$= \left[ \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dfrac{4}{81}x^3 \right]_{0}^{1.5}$

Using $1.5 = \dfrac{3}{2}$:

$= \left( \dfrac{1}{3}\left(\dfrac{3}{2}\right) - \dfrac{1}{9}\left(\dfrac{3}{2}\right)^2 + \dfrac{4}{81}\left(\dfrac{3}{2}\right)^3 \right) - (0)$

$= \dfrac{1}{2} - \dfrac{1}{9}\left(\dfrac{9}{4}\right) + \dfrac{4}{81}\left(\dfrac{27}{8}\right)$

$= \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{6}$

$= \dfrac{6}{12} - \dfrac{3}{12} + \dfrac{2}{12}$

$= \dfrac{5}{12}$