已知 $f(x)=\dfrac{1}{4+9x^2}$。

1. Factor out the coefficient of $x^2$ in the denominator to match the standard integral form $\frac{1}{a^2+x^2}$.

$\int \frac{1}{4+9x^2} \,dx = \frac{1}{9} \int \frac{1}{\frac{4}{9}+x^2} \,dx$

2. Integrate using the standard inverse tangent formula, remembering to add the constant of integration.

$= \frac{1}{9} \left[ \frac{1}{\frac{2}{3}} \tan^{-1}\left(\frac{x}{\frac{2}{3}}\right) \right] + C \n= \frac{1}{9} \left( \frac{3}{2} \tan^{-1}\left(\frac{3x}{2}\right) \right) + C \n= \frac{1}{6} \tan^{-1}\left(\frac{3x}{2}\right) + C$

1. Rewrite the function into the standard form $A(1+y)^n$ suitable for binomial expansion.

$f(x) = (4+9x^2)^{-1} = \left[4\left(1+\dfrac{9}{4}x^2\right)\right]^{-1} = \dfrac{1}{4}\left(1+\dfrac{9}{4}x^2\right)^{-1}$

2. Apply the standard binomial expansion formula $(1+y)^n$ up to the relevant term.

$\dfrac{1}{4}\left(1+\dfrac{9}{4}x^2\right)^{-1} = \dfrac{1}{4} \left[ 1 + (-1)\left(\dfrac{9}{4}x^2\right) + \dfrac{(-1)(-2)}{2!} \left(\dfrac{9}{4}x^2\right)^2 + \dots \right]$

3. Simplify the terms and distribute the constant factor to find the final coefficients.

$= \dfrac{1}{4} \left( 1 - \dfrac{9}{4}x^2 + \dfrac{81}{16}x^4 \right)$

$= \dfrac{1}{4} - \dfrac{9}{16}x^2 + \dfrac{81}{64}x^4$

1. Evaluate the results from parts (i) and (ii) concisely, as they are required for the 'Hence' method.

From (i), $\int f(x)\,dx = \int \frac{1}{4+9x^2} \,dx = \frac{1}{2 \times 3}\tan^{-1}\left(\frac{3x}{2}\right) = \frac{1}{6}\tan^{-1}\left(\frac{3x}{2}\right)$.
From (ii), the binomial expansion of $f(x)$ is:
$f(x) = \frac{1}{4}\left(1 + \frac{9x^2}{4}\right)^{-1}$
$= \frac{1}{4}\left[1 - \left(\frac{9x^2}{4}\right) + \left(\frac{9x^2}{4}\right)^2 - \dots\right]$
$= \frac{1}{4} - \frac{9}{16}x^2 + \frac{81}{64}x^4 - \dots$

2. Express the inverse tangent function in terms of an integral of the series.

Using the integral from (i):
$\tan^{-1}\left(\frac{3x}{2}\right) = 6 \int f(x) \,dx$
Substituting the expansion from (ii):
$\tan^{-1}\left(\frac{3x}{2}\right) = 6 \int \left( \frac{1}{4} - \frac{9}{16}x^2 + \frac{81}{64}x^4 \right) \,dx$

3. Integrate the series term-by-term.

$\tan^{-1}\left(\frac{3x}{2}\right) = 6 \left[ \frac{1}{4}x - \frac{9}{16}\left(\frac{x^3}{3}\right) + \frac{81}{64}\left(\frac{x^5}{5}\right) \right] + C$
$= 6 \left( \frac{1}{4}x - \frac{3}{16}x^3 + \frac{81}{320}x^5 \right) + C$
$= \frac{3}{2}x - \frac{9}{8}x^3 + \frac{243}{160}x^5 + C$

4. Evaluate the constant of integration to find the final series.

Substitute $x = 0$ into the series equation:
$\tan^{-1}(0) = \frac{3}{2}(0) - \frac{9}{8}(0)^3 + \frac{243}{160}(0)^5 + C$
$0 = 0 + C \implies C = 0$
Therefore, the Maclaurin series for $\tan^{-1}\left(\frac{3x}{2}\right)$ is:
$\frac{3}{2}x - \frac{9}{8}x^3 + \frac{243}{160}x^5 + \dots$

1. State the Maclaurin series for $\tan^{-1}\frac{3x}{2}$ derived in part (iii).

From part (iii), the Maclaurin series up to the term in $x^5$ is:
$\tan^{-1}\left(\frac{3x}{2}\right) \approx \frac{3}{2}x - \frac{9}{8}x^3 + \frac{243}{160}x^5$

2. Substitute the series into the definite integral and integrate term-by-term.

$\int_{0}^{0.5}\tan^{-1}\frac{3x}{2}\,dx \approx \int_{0}^{0.5} \left( \frac{3}{2}x - \frac{9}{8}x^3 + \frac{243}{160}x^5 \right) \,dx$
$= \left[ \frac{3}{4}x^2 - \frac{9}{32}x^4 + \frac{81}{320}x^6 \right]_{0}^{0.5}$

3. Evaluate the definite integral by substituting the limits.

$= \left( \frac{3}{4}(0.5)^2 - \frac{9}{32}(0.5)^4 + \frac{81}{320}(0.5)^6 \right) - 0$
$= \frac{3}{16} - \frac{9}{512} + \frac{81}{20480}$
$= 0.173876...$
$\approx 0.174 \text{ (correct to 3 d.p.)}$

1. Evaluate the definite integral using a graphing calculator

$\int_{0}^{0.5}\tan^{-1}\dfrac{3x}{2}\,dx \approx 0.172988... \approx 0.173 \text{ (to 3 d.p.)}$

1. Evaluate the accuracy by comparing the estimate from part (iv) to the exact value from part (v).

Based on the earlier parts, the series estimate in (iv) is $\approx 0.174$ and the exact calculator value in (v) is $\approx 0.173$.

The estimate in (iv) is very close to the true value and is accurate to 2 decimal places.

2. Suggest a method to improve the accuracy of the series approximation.

The accuracy of the estimate can be improved by including more terms (higher powers of $x$) in the Maclaurin series expansion.