By means of the substitution $x = \operatorname{cosec}\theta$, show that $\int_{\frac{2}{\sqrt{3}}}^{2} \frac{1}{x^3 \sqrt{x^2 - 1}} \mathrm{d}x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2(\theta) \mathrm{d}\theta$. You are to show all workings clearly. Hence evaluate exactly $\int_{\frac{2}{\sqrt{3}}}^{2} \frac{1}{x^3 \sqrt{x^2 - 1}} \mathrm{d}x$.

1. Differentiate the given substitution $x = \operatorname{cosec}\theta$ to find an expression for $\mathrm{d}x$.

$x = \operatorname{cosec}\theta$\n$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -\operatorname{cosec}\theta \cot\theta$\n$\mathrm{d}x = -\operatorname{cosec}\theta \cot\theta \,\mathrm{d}\theta$

2. Determine the new upper and lower limits of integration in terms of $\theta$.

When $x = \frac{2}{\sqrt{3}}$:\n$\operatorname{cosec}\theta = \frac{2}{\sqrt{3}} \implies \sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$\n\nWhen $x = 2$:\n$\operatorname{cosec}\theta = 2 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$

3. Substitute $x$, $\mathrm{d}x$, and the new limits into the integral, and apply the trigonometric identity $\operatorname{cosec}^2\theta - 1 = \cot^2\theta$.

$\begin{aligned} \int_{\frac{2}{\sqrt{3}}}^{2} \frac{1}{x^3 \sqrt{x^2 - 1}} \,\mathrm{d}x &= \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{1}{\operatorname{cosec}^3\theta \sqrt{\operatorname{cosec}^2\theta - 1}} (-\operatorname{cosec}\theta \cot\theta) \,\mathrm{d}\theta \\ &= \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{-\operatorname{cosec}\theta \cot\theta}{\operatorname{cosec}^3\theta \sqrt{\cot^2\theta}} \,\mathrm{d}\theta \end{aligned}$\n*(Note: For $\theta \in [\frac{\pi}{6}, \frac{\pi}{3}]$, $\cot\theta > 0$, so $\sqrt{\cot^2\theta} = \cot\theta$)*\n$\int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{-\operatorname{cosec}\theta \cot\theta}{\operatorname{cosec}^3\theta \cot\theta} \,\mathrm{d}\theta$

4. Simplify the integrand and invert the limits to absorb the negative sign.

$\begin{aligned} \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \frac{-1}{\operatorname{cosec}^2\theta} \,\mathrm{d}\theta &= \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} -\sin^2\theta \,\mathrm{d}\theta \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2\theta \,\mathrm{d}\theta \quad \text{(shown)} \end{aligned}$

1. Apply the result from part (i) and rewrite the integrand using the double angle identity.

$\int_{\frac{2}{\sqrt{3}}}^{2} \frac{1}{x^3 \sqrt{x^2 - 1}} \mathrm{d}x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2(\theta) \mathrm{d}\theta$
$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 - \cos(2\theta)}{2} \mathrm{d}\theta$

2. Integrate the trigonometric expression.

$= \frac{1}{2} \left[ \theta - \frac{1}{2}\sin(2\theta) \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

3. Substitute the limits of integration to evaluate the exact value.

$= \frac{1}{2} \left( \frac{\pi}{3} - \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) \right) - \frac{1}{2} \left( \frac{\pi}{6} - \frac{1}{2}\sin\left(\frac{\pi}{3}\right) \right)$
$= \frac{1}{2} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) - \frac{1}{2} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)$
$= \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right) - \left( \frac{\pi}{12} - \frac{\sqrt{3}}{8} \right)$
$= \frac{\pi}{6} - \frac{\pi}{12}$
$= \frac{\pi}{12}$