在本题中，你可以使用公式表（MF27）中的展开式。

1. Substitute the standard Maclaurin expansion for $\cos 3x$.

$\cos 3x \approx 1 - \frac{(3x)^2}{2!} + \frac{(3x)^4}{4!} = 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4$

2. Substitute into the original expression and factorise to prepare for the logarithmic expansion.

$\ln(1 + \cos 3x) \approx \ln\left(1 + 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4\right)$
$= \ln\left(2 - \frac{9}{2}x^2 + \frac{27}{8}x^4\right)$
$= \ln\left[2\left(1 - \frac{9}{4}x^2 + \frac{27}{16}x^4\right)\right]$
$= \ln 2 + \ln\left(1 - \frac{9}{4}x^2 + \frac{27}{16}x^4\right)$

3. Use the standard expansion for $\ln(1 + u)$ where $u = -\frac{9}{4}x^2 + \frac{27}{16}x^4$.

$\ln(1 + u) \approx u - \frac{u^2}{2}$
$= \left(- \frac{9}{4}x^2 + \frac{27}{16}x^4\right) - \frac{1}{2}\left(- \frac{9}{4}x^2 + \frac{27}{16}x^4\right)^2$
$= - \frac{9}{4}x^2 + \frac{27}{16}x^4 - \frac{1}{2}\left(\frac{81}{16}x^4\right) + \dots$

4. Simplify the expansion up to $x^4$.

$= - \frac{9}{4}x^2 + \left(\frac{27}{16} - \frac{81}{32}\right)x^4$
$= - \frac{9}{4}x^2 + \left(\frac{54}{32} - \frac{81}{32}\right)x^4$
$= - \frac{9}{4}x^2 - \frac{27}{32}x^4$
Therefore, the expansion is:
$\ln 2 - \frac{9}{4}x^2 - \frac{27}{32}x^4$

1. Derive the Maclaurin expansion of $\ln(1 + \cos 3x)$ up to the term in $x^4$ (as required from part (i)).

From part (i), we require the expansion of $\ln(1 + \cos 3x)$:
Using the standard Maclaurin series for $\cos x$:
$\cos 3x \approx 1 - \frac{(3x)^2}{2!} + \frac{(3x)^4}{4!} = 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4$
Substitute this into the natural logarithm:
$\ln(1 + \cos 3x) \approx \ln\left(2 - \frac{9}{2}x^2 + \frac{27}{8}x^4\right)$
$= \ln\left[ 2 \left( 1 - \frac{9}{4}x^2 + \frac{27}{16}x^4 \right) \right]$
$= \ln 2 + \ln\left(1 + \left(-\frac{9}{4}x^2 + \frac{27}{16}x^4\right)\right)$
Using the standard series $\ln(1+y) \approx y - \frac{y^2}{2}$:
$\approx \ln 2 + \left(-\frac{9}{4}x^2 + \frac{27}{16}x^4\right) - \frac{1}{2}\left(-\frac{9}{4}x^2\right)^2$
$= \ln 2 - \frac{9}{4}x^2 + \frac{27}{16}x^4 - \frac{81}{32}x^4$
$= \ln 2 - \frac{9}{4}x^2 - \frac{27}{32}x^4$

2. Substitute the expansion into the integral and integrate term by term.

Substitute the expansion into the given integral:
$\int_{0}^{0.5} x \ln(1 + \cos 3x) \, \text{d}x \approx \int_{0}^{0.5} x \left( \ln 2 - \frac{9}{4}x^2 - \frac{27}{32}x^4 \right) \, \text{d}x$
$= \int_{0}^{0.5} \left( x \ln 2 - \frac{9}{4}x^3 - \frac{27}{32}x^5 \right) \, \text{d}x$
$= \left[ \frac{x^2}{2} \ln 2 - \frac{9}{16}x^4 - \frac{27}{192}x^6 \right]_{0}^{0.5}$
$= \left[ \frac{x^2}{2} \ln 2 - \frac{9}{16}x^4 - \frac{9}{64}x^6 \right]_{0}^{0.5}$

3. Evaluate the definite integral limits and round to 5 decimal places.

Substitute the limits $0$ and $0.5$:
$= \left( \frac{(0.5)^2}{2} \ln 2 - \frac{9}{16}(0.5)^4 - \frac{9}{64}(0.5)^6 \right) - 0$
$= \frac{1}{8} \ln 2 - \frac{9}{256} - \frac{9}{4096}$
$\approx 0.0866434 - 0.0351563 - 0.0021973$
$= 0.04928988...$
Rounding to 5 decimal places:
$\approx 0.04929$

1. Evaluate the definite integral using a graphing calculator

Using a graphing calculator,
$\int_{0}^{0.5} x \ln(1 + \cos 3x) \, \text{d}x \approx 0.04900 \text{ (to 5 d.p.)}$

1. Sketch the suitable diagram showing both the exact and approximated curves.

Let $y_1 = x \ln(1+\cos 3x)$ (exact curve) and $y_2 = x \left(\ln 2 - \frac{9}{4}x^2 - \frac{27}{32}x^4\right)$ (approximation curve used in part (ii)).
Using a graphing calculator, sketch both graphs for $0 \le x \le 0.5$.
(Self-drawn diagram showing both curves starting at the origin. For $x > 0$, $y_2$ lies above $y_1$, and the gap widens as $x$ approaches $0.5$.)

2. Comment on the accuracy by relating the diagram to the answers in (ii) and (iii).

From parts (ii) and (iii), the approximate integral value (approx $0.04929$) is larger than the exact integral value (approx $0.04533$).
The diagram visually confirms this because the approximation curve $y_2$ lies above the exact curve $y_1$ on the interval $0 < x \le 0.5$, making the area under $y_2$ an overestimate.
The Maclaurin expansion is an approximation centred at $x=0$, so it is highly accurate near $x=0$. However, as $x$ increases towards $0.5$, the approximation curve diverges significantly from the actual function, causing the reduced accuracy observed.