Using the substitution $u = \cos x$, find $\displaystyle \int \frac{\sin 2x}{\cos^2 x + \cos x + 2}\,dx$, giving your answer in terms of $x$.

1. Find the derivative of the substitution variable and apply the double angle formula for sine.

$u = \cos x \implies \frac{du}{dx} = -\sin x \implies dx = -\frac{1}{\sin x} \, du$
$\sin 2x = 2 \sin x \cos x$

2. Substitute $u$ into the integral to eliminate $x$.

$\int \frac{2 \sin x \cos x}{\cos^2 x + \cos x + 2} \, dx = \int \frac{2 \sin x (u)}{u^2 + u + 2} \left( -\frac{1}{\sin x} \right) \, du$
$= \int \frac{-2u}{u^2 + u + 2} \, du$

3. Manipulate the numerator to extract the exact derivative of the denominator.

The derivative of $u^2 + u + 2$ is $2u + 1$.
$\int \frac{-2u}{u^2 + u + 2} \, du = -\int \frac{2u + 1 - 1}{u^2 + u + 2} \, du$
$= -\int \frac{2u + 1}{u^2 + u + 2} \, du + \int \frac{1}{u^2 + u + 2} \, du$

4. Complete the square for the denominator of the second integral.

$u^2 + u + 2 = \left( u + \frac{1}{2} \right)^2 - \frac{1}{4} + 2$
$= \left( u + \frac{1}{2} \right)^2 + \frac{7}{4}$

5. Integrate using standard forms and substitute back $u = \cos x$.

$= -\ln(u^2 + u + 2) + \frac{1}{\frac{\sqrt{7}}{2}} \tan^{-1} \left( \frac{u + \frac{1}{2}}{\frac{\sqrt{7}}{2}} \right) + C$
$= -\ln(\cos^2 x + \cos x + 2) + \frac{2}{\sqrt{7}} \tan^{-1} \left( \frac{2\cos x + 1}{\sqrt{7}} \right) + C$