It is given that $\mathrm{f}(x) = \ln(a + x)$, $x \in \mathbb{R}$, $x > -a$, where $a$ is a constant.

1. Factorise $a$ from the argument of the logarithm to obtain the form $1+u$, allowing the use of the standard series.

$\mathrm{f}(x) = \ln(a+x) = \ln\left[a\left(1+\frac{x}{a}\right)\right] = \ln a + \ln\left(1+\frac{x}{a}\right)$

2. Substitute $u = \frac{x}{a}$ into the standard series expansion for $\ln(1+x)$ up to the term in $x^3$ and simplify.

$\begin{aligned} \mathrm{f}(x) &\approx \ln a + \left(\frac{x}{a}\right) - \frac{1}{2}\left(\frac{x}{a}\right)^2 + \frac{1}{3}\left(\frac{x}{a}\right)^3 \\ &= \ln a + \frac{x}{a} - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} \end{aligned}$

1. State the standard series for $\ln(1+x)$ from part (a).

Given $a = 1$, $\mathrm{f}(x) = \ln(1+x)$.
$\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + \dots$

2. Substitute the series for $\ln(1+x)$ into the standard series for $\sin u$.

Using the standard series $\sin u = u - \frac{u^3}{3!} + \dots$, substitute $u = \ln(1+x)$:
$\sin(\ln(1+x)) = \left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right) - \frac{1}{6}\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)^3 + \dots$

3. Expand and simplify the terms up to $x^3$.

In the expansion of $\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)^3$, the only term with a power up to $x^3$ is $x^3$.
$\sin(\ln(1+x)) \approx x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{6}(x^3)$
$= x - \frac{1}{2}x^2 + \left(\frac{1}{3} - \frac{1}{6}\right)x^3$
$= x - \frac{1}{2}x^2 + \frac{1}{6}x^3$ (Shown)

1. Differentiate the series for $\sin[\mathrm{f}(x)]$ from part (b) with respect to $x$, applying the chain rule.

From part (b), we have:
$\sin[\mathrm{f}(x)] \approx x - \frac{1}{2}x^2 + \frac{1}{6}x^3$

Differentiating both sides with respect to $x$:
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin[\mathrm{f}(x)]\right) \approx \frac{\mathrm{d}}{\mathrm{d}x}\left(x - \frac{1}{2}x^2 + \frac{1}{6}x^3\right)$
$\cos[\mathrm{f}(x)] \cdot \mathrm{f}'(x) \approx 1 - x + \frac{1}{2}x^2$

Given $\mathrm{f}(x) = \ln(1+x)$ from substituting $a=1$, its derivative is $\mathrm{f}'(x) = \frac{1}{1+x}$. Substituting this in gives:
$\cos[\mathrm{f}(x)] \left(\frac{1}{1+x}\right) \approx 1 - x + \frac{1}{2}x^2$

2. Multiply by $(1+x)$ and expand to deduce the series up to $x^2$.

$\cos[\mathrm{f}(x)] \approx (1+x)\left(1 - x + \frac{1}{2}x^2\right)$
Expanding the expression and ignoring terms with powers higher than $x^2$:
$\cos[\mathrm{f}(x)] \approx 1 - x + \frac{1}{2}x^2 + x(1) - x(x)$
$\cos[\mathrm{f}(x)] \approx 1 - x + \frac{1}{2}x^2 + x - x^2$
$\cos[\mathrm{f}(x)] \approx 1 - \frac{1}{2}x^2$

1. Evaluate the definite integral of the given polynomial.

2. Explain whether the value is a good approximation to the integral of the original function.

From part (b), $x - \frac{1}{2}x^2 + \frac{1}{6}x^3$ is the Maclaurin series expansion of $\sin[\mathrm{f}(x)]$.

A Maclaurin series is an expansion about $x = 0$ and is only a good approximation for small values of $x$ (i.e. close to $0$).

Since the limits of integration are from $x = 1$ to $x = 3$, which are not close to $0$, this value is not a good approximation.