Find the exact value of the following integrals.

1. Apply integration by substitution by letting $u = \sin^{-1} x$.

Let $u = \sin^{-1} x$.
Then $\frac{du}{dx} = \frac{1}{\sqrt{1-x^2}}$, which gives $du = \frac{1}{\sqrt{1-x^2}} \, dx$.

2. Determine the new limits of integration in terms of $u$.

When $x = \frac{\sqrt{2}}{2}$, $u = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$.
When $x = \frac{\sqrt{3}}{2}$, $u = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$.

3. Substitute into the integral and evaluate to find the exact value.

$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 2u \, du = [u^2]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$
$= \left(\frac{\pi}{3}\right)^2 - \left(\frac{\pi}{4}\right)^2$
$= \frac{\pi^2}{9} - \frac{\pi^2}{16}$
$= \frac{16\pi^2 - 9\pi^2}{144}$
$= \frac{7\pi^2}{144}$

1. Find where the expression inside the modulus changes sign in the given interval.

$\cos 2x = 0$
$2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}$
Since $0 \leq x \leq \frac{\pi}{3}$, the sign changes at $x = \frac{\pi}{4}$.

2. Split the definite integral to resolve the modulus.

For $0 \leq x \leq \frac{\pi}{4}$, $\cos 2x \geq 0$.
For $\frac{\pi}{4} < x \leq \frac{\pi}{3}$, $\cos 2x < 0$.
$\int_{0}^{\frac{\pi}{3}} |\cos 2x|\,dx = \int_{0}^{\frac{\pi}{4}} \cos 2x \,dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (-\cos 2x) \,dx$

3. Integrate and evaluate using the exact trigonometric values.

$= \left[ \frac{1}{2} \sin 2x \right]_0^{\frac{\pi}{4}} - \left[ \frac{1}{2} \sin 2x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$
$= \frac{1}{2}\left( \sin\frac{\pi}{2} - \sin 0 \right) - \frac{1}{2}\left( \sin\frac{2\pi}{3} - \sin\frac{\pi}{2} \right)$
$= \frac{1}{2}(1 - 0) - \frac{1}{2}\left( \frac{\sqrt{3}}{2} - 1 \right)$
$= \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{1}{2}$
$= 1 - \frac{\sqrt{3}}{4}$

1. Express the quadratic denominator in the completed square form.

$-x^2 + 2kx + 3k^2 = -(x^2 - 2kx) + 3k^2$
$= -\left((x-k)^2 - k^2\right) + 3k^2$
$= 4k^2 - (x-k)^2$
$= (2k)^2 - (x-k)^2$

2. Evaluate the integral using the standard formula for $\frac{1}{a^2-x^2}$.

$\int \frac{1}{-x^2+2kx+3k^2}\,dx = \int \frac{1}{(2k)^2 - (x-k)^2}\,dx$
$= \frac{1}{2(2k)} \ln \left| \frac{2k + (x-k)}{2k - (x-k)} \right| + C$
$= \frac{1}{4k} \ln \left| \frac{x+k}{3k-x} \right| + C$