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

1. Rewrite $f(x)$ to isolate a $1$ inside the logarithm, matching the form $\ln(1+y)$.

$\begin{aligned} f(x) &= \ln(a+x) \\ &= \ln\left[a\left(1+\frac{x}{a}\right)\right] \\ &= \ln a + \ln\left(1+\frac{x}{a}\right) \end{aligned}$

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

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

1. State the series expansion of $f(x)$ for $a=1$ and substitute it into the standard Maclaurin series for $\sin x$.

When $a=1$, $f(x) = \ln(1+x)$.
Using the standard series for $\ln(1+x)$, we have $f(x) \approx x - \frac{1}{2}x^2 + \frac{1}{3}x^3$.
Using the standard series $\sin u \approx u - \frac{1}{6}u^3$, we substitute $u = f(x)$:
$\sin\bigl[f(x)\bigr] \approx \left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right) - \frac{1}{6}\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)^3$

2. Expand the expression and collect terms up to $x^3$.

Expanding and keeping only terms up to $x^3$, we note that the lowest degree term of $\left(x - \frac{1}{2}x^2 + \frac{1}{3}x^3\right)^3$ is $x^3$. Higher powers are ignored.
$\sin\bigl[f(x)\bigr] \approx x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{6}(x^3)$
$= x - \frac{1}{2}x^2 + \left(\frac{1}{3} - \frac{1}{6}\right)x^3$
$= x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \quad \text{(shown)}$

1. Differentiate the Maclaurin series of $\sin\bigl[f(x)\bigr]$ from part (b) to establish a relationship with $\cos\bigl[f(x)\bigr]$.

$\frac{\mathrm{d}}{\mathrm{d}x} \Bigl(\sin\bigl[f(x)\bigr]\Bigr) = \cos\bigl[f(x)\bigr] \cdot f'(x)$
Since $a=1$, $f(x) = \ln(1+x)$ and $f'(x) = \frac{1}{1+x}$.
Using the series from part (b):
$\cos\bigl[f(x)\bigr] \cdot \Bigl(\frac{1}{1+x}\Bigr) \approx \frac{\mathrm{d}}{\mathrm{d}x} \Bigl(x - \frac{1}{2}x^{2} + \frac{1}{6}x^{3}\Bigr)$
$\frac{\cos\bigl[f(x)\bigr]}{1+x} \approx 1 - x + \frac{1}{2}x^{2}$

2. Rearrange and expand the expression to find the series for $\cos\bigl[f(x)\bigr]$ up to the term in $x^2$.

$\cos\bigl[f(x)\bigr] \approx (1+x)\Bigl(1 - x + \frac{1}{2}x^{2}\Bigr)$
$\cos\bigl[f(x)\bigr] \approx 1 - x + \frac{1}{2}x^{2} + x - x^{2} + \frac{1}{2}x^{3}$
Ignoring terms involving powers of $x^3$ and higher:
$\cos\bigl[f(x)\bigr] \approx 1 - \frac{1}{2}x^{2}$

1. Evaluate the given definite integral of the polynomial.

$\int_{1}^{3} \Bigl(x - \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3\Bigr) \,\mathrm{d}x = \left[ \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4 \right]_1^3$
$= \left( \frac{1}{2}(9) - \frac{1}{6}(27) + \frac{1}{24}(81) \right) - \left( \frac{1}{2}(1) - \frac{1}{6}(1) + \frac{1}{24}(1) \right)$
$= \left( \frac{9}{2} - \frac{9}{2} + \frac{27}{8} \right) - \left( \frac{12}{24} - \frac{4}{24} + \frac{1}{24} \right)$
$= \frac{27}{8} - \frac{9}{24}$
$= \frac{27}{8} - \frac{3}{8} = \frac{24}{8} = 3$

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

The series expansion $x-\tfrac{1}{2}x^{2}+\tfrac{1}{6}x^{3}$ is a Maclaurin series approximation of $\sin\bigl[f(x)\bigr]$.

By definition, a Maclaurin series is an expansion about $x = 0$ and truncating it gives a polynomial approximation that is most accurate near $x = 0$.

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