Find $\int \cot^2 3x \; \mathrm{d}x$. Find the exact value of $\int_{2}^{4} \frac{x^2-4x+3}{x^2-4x+8} \; \mathrm{d}x$. Use the substitution $x = 2\cot\theta$ to find the exact value of $\int_{\frac{2\sqrt{3}}{3}}^{2} \frac{4-x^2}{(4+x^2)^2} \; \mathrm{d}x$.

1. Apply the trigonometric identity $\cot^2 \theta = \csc^2 \theta - 1$ to rewrite the integrand.

$\int \cot^2 3x \; \mathrm{d}x = \int (\csc^2 3x - 1) \; \mathrm{d}x$

2. Integrate term-by-term using the standard integral of $\csc^2(ax)$.

$= -\frac{1}{3}\cot 3x - x + c$

1. Rewrite the improper rational fraction by splitting the numerator, and complete the square for the denominator.

$\frac{x^2-4x+3}{x^2-4x+8} = \frac{(x^2-4x+8) - 5}{x^2-4x+8} = 1 - \frac{5}{x^2-4x+8}$
$x^2-4x+8 = (x-2)^2 - 4 + 8 = (x-2)^2 + 4$

2. Integrate the function using the standard formula for $\frac{1}{x^2+a^2}$.

$\int_{2}^{4} \left( 1 - \frac{5}{(x-2)^2+2^2} \right) \; \mathrm{d}x = \left[ x - \frac{5}{2} \tan^{-1}\left(\frac{x-2}{2}\right) \right]_{2}^{4}$

3. Substitute the limits to evaluate the definite integral exactly.

$= \left( 4 - \frac{5}{2}\tan^{-1}\left(\frac{4-2}{2}\right) \right) - \left( 2 - \frac{5}{2}\tan^{-1}\left(\frac{2-2}{2}\right) \right)$
$= \left( 4 - \frac{5}{2}\tan^{-1}(1) \right) - \left( 2 - \frac{5}{2}\tan^{-1}(0) \right)$
$= 4 - \frac{5}{2}\left(\frac{\pi}{4}\right) - 2$
$= 2 - \frac{5\pi}{8}$

1. Find the differential $\mathrm{d}x$ and the new limits of integration in terms of $\theta$.

$x = 2\cot\theta \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\csc^2\theta \implies \mathrm{d}x = -2\csc^2\theta \; \mathrm{d}\theta$

Find the new limits:
When $x = \frac{2\sqrt{3}}{3}$, $2\cot\theta = \frac{2\sqrt{3}}{3} \implies \tan\theta = \sqrt{3} \implies \theta = \frac{\pi}{3}$
When $x = 2$, $2\cot\theta = 2 \implies \tan\theta = 1 \implies \theta = \frac{\pi}{4}$

2. Substitute $x$, $\mathrm{d}x$, and the limits into the integral.

$\int_{\frac{2\sqrt{3}}{3}}^{2} \frac{4-x^2}{(4+x^2)^2} \; \mathrm{d}x = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{4 - 4\cot^2\theta}{(4 + 4\cot^2\theta)^2} (-2\csc^2\theta) \; \mathrm{d}\theta$

$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{4(1 - \cot^2\theta)}{(4\csc^2\theta)^2} (-2\csc^2\theta) \; \mathrm{d}\theta$

3. Simplify the integrand using trigonometric identities.

$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{4(1 - \cot^2\theta)}{16\csc^4\theta} (-2\csc^2\theta) \; \mathrm{d}\theta$

$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{-8(1 - \cot^2\theta)\csc^2\theta}{16\csc^4\theta} \; \mathrm{d}\theta$

$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} -\frac{1}{2} \left(\frac{1 - \cot^2\theta}{\csc^2\theta}\right) \; \mathrm{d}\theta$

Using $\frac{1}{\csc^2\theta} = \sin^2\theta$ and $\frac{\cot^2\theta}{\csc^2\theta} = \cos^2\theta$:
$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} -\frac{1}{2} (\sin^2\theta - \cos^2\theta) \; \mathrm{d}\theta$

$= \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{1}{2} \cos 2\theta \; \mathrm{d}\theta$

4. Integrate and evaluate to find the exact value.

$= \left[ \frac{1}{4} \sin 2\theta \right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}$

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

$= \frac{1}{4} \left( 1 - \frac{\sqrt{3}}{2} \right)$

$= \frac{2 - \sqrt{3}}{8}$