Using standard series, find the Maclaurin series expansion of $\dfrac{1}{\sqrt{1 - a^2 x^2}}$ up to and including the term in $x^4$, where $a$ is a positive constant. Find also the range of values of $x$, in terms of $a$, for which the expansion is valid.

1. Rewrite the given expression in the index form suitable for binomial expansion.

$\dfrac{1}{\sqrt{1 - a^2 x^2}} = (1 - a^2 x^2)^{-\frac{1}{2}}$

2. Apply the standard binomial series expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$ with $y = -a^2x^2$ and $n = -\frac{1}{2}$.

$(1 - a^2 x^2)^{-\frac{1}{2}} = 1 + \left(-\dfrac{1}{2}\right)(-a^2 x^2) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} (-a^2 x^2)^2 + \dots$

3. Simplify the coefficients to obtain the expansion up to $x^4$.

$= 1 + \dfrac{1}{2} a^2 x^2 + \dfrac{\frac{3}{4}}{2} a^4 x^4 + \dots$
$= 1 + \dfrac{1}{2} a^2 x^2 + \dfrac{3}{8} a^4 x^4 + \dots$

4. Identify the condition for convergence of the binomial series.

For the expansion to be valid, we need $|-a^2 x^2| < 1$.
$a^2 x^2 < 1$

5. Solve the inequality to find the range of values for $x$.

$|ax| < 1$
Since $a$ is a positive constant, $a|x| < 1$.
$|x| < \dfrac{1}{a}$
$\therefore -\dfrac{1}{a} < x < \dfrac{1}{a}$

1. Concisely evaluate the Maclaurin series expansion of $\dfrac{1}{\sqrt{1 - a^2 x^2}}$ as required from part (a).

From part (a), using the standard binomial expansion for $(1 - a^2 x^2)^{-\frac{1}{2}}$:
$(1 - a^2 x^2)^{-\frac{1}{2}} \approx 1 + \left(-\frac{1}{2}\right)(-a^2 x^2) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} (-a^2 x^2)^2$
$= 1 + \frac{1}{2}a^2 x^2 + \frac{3}{8}a^4 x^4$

2. Substitute a suitable value for $a$ and establish the relationship with the derivative of $\cos^{-1} x$.

By choosing the positive constant $a = 1$:
$\frac{1}{\sqrt{1 - x^2}} \approx 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4$
Using the standard differentiation result $\frac{\mathrm{d}}{\mathrm{d}x}(\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}}$:
$\frac{\mathrm{d}}{\mathrm{d}x}(\cos^{-1} x) \approx -1 - \frac{1}{2}x^2 - \frac{3}{8}x^4$

3. Integrate the series term-by-term and evaluate the constant of integration to complete the proof.

Integrating both sides with respect to $x$:
$\cos^{-1} x \approx \int \left( -1 - \frac{1}{2}x^2 - \frac{3}{8}x^4 \right) \, \mathrm{d}x$
$\cos^{-1} x \approx C - x - \frac{1}{6}x^3 - \frac{3}{40}x^5$
To find the constant of integration $C$, substitute $x = 0$:
$\cos^{-1}(0) = C - 0 \implies C = \frac{\pi}{2}$
$\therefore \cos^{-1} x \approx \frac{\pi}{2} - x - \frac{1}{6}x^3 - \frac{3}{40}x^5 \text{ (shown)}$

1. Substitute the polynomial approximation of $\cos^{-1} x$ into the definite integral.

$\int_{0}^{\frac{1}{2}} \cos^{-1} x\, dx \approx \int_{0}^{\frac{1}{2}} \left( \frac{\pi}{2} - x - \frac{1}{6} x^3 - \frac{3}{40} x^5 \right) dx$

2. Integrate the polynomial term-by-term.

$= \left[ \frac{\pi}{2} x - \frac{x^2}{2} - \frac{x^4}{24} - \frac{3x^6}{240} \right]_{0}^{\frac{1}{2}}$
$= \left[ \frac{\pi}{2} x - \frac{x^2}{2} - \frac{x^4}{24} - \frac{x^6}{80} \right]_{0}^{\frac{1}{2}}$

3. Evaluate the definite integral using the limits and compute the final answer to 5 significant figures.

$= \left( \frac{\pi}{2}\left(\frac{1}{2}\right) - \frac{1}{2}\left(\frac{1}{2}\right)^2 - \frac{1}{24}\left(\frac{1}{2}\right)^4 - \frac{1}{80}\left(\frac{1}{2}\right)^6 \right) - 0$
$= \frac{\pi}{4} - \frac{1}{8} - \frac{1}{384} - \frac{1}{5120}$
$= 0.65759868...$
$\approx 0.65760 \text{ (to 5 s.f.)}$

1. Evaluate the actual value using a graphing calculator and comment on the accuracy of the estimate

Using a graphing calculator, the actual value of the integral is:
$\displaystyle \int_{0}^{\tfrac{1}{2}} \cos^{-1} x\, dx \approx 0.65757$ (to 5 s.f.)

The estimate calculated in (c) is $0.65760$.
Comparing the two, the difference is only $0.00003$. Therefore, the estimate is highly accurate.

2. Suggest a method to improve the approximation

The estimate can be further improved by including more terms (i.e., higher powers of $x$) in the Maclaurin series expansion of $\cos^{-1} x$.