Given that $y = 2\cos x \sin 2x - \sin x \cos 2x$, show that $\dfrac{dy}{dx} = k\cos x \cos 2x$, stating the value of $k$. Hence, show that $\int_{0}^{\pi/4}\left(7\cos x\cos 2x + \sec^{2} x\right)\,dx = \dfrac{7\sqrt{2}}{3} + 1.$

1. Differentiate the first term $2\cos x \sin 2x$ using the product rule.

Let $u_1 = 2\cos x$ and $v_1 = \sin 2x$.
$\frac{d}{dx}(2\cos x \sin 2x) = 2(-\sin x)(\sin 2x) + (2\cos x)(2\cos 2x)$
$= -2\sin x \sin 2x + 4\cos x \cos 2x$

2. Differentiate the second term $\sin x \cos 2x$ using the product rule.

Let $u_2 = \sin x$ and $v_2 = \cos 2x$.
$\frac{d}{dx}(\sin x \cos 2x) = (\cos x)(\cos 2x) + (\sin x)(-2\sin 2x)$
$= \cos x \cos 2x - 2\sin x \sin 2x$

3. Combine the derivatives and simplify to find $k$.

$\frac{dy}{dx} = (-2\sin x \sin 2x + 4\cos x \cos 2x) - (\cos x \cos 2x - 2\sin x \sin 2x)$
$\frac{dy}{dx} = -2\sin x \sin 2x + 4\cos x \cos 2x - \cos x \cos 2x + 2\sin x \sin 2x$
$\frac{dy}{dx} = 3\cos x \cos 2x$
Comparing with $\frac{dy}{dx} = k\cos x \cos 2x$, we have $k = 3$.

1. Determine the value of $k$ from part (a) to use the reverse differentiation property.

From part (a), differentiating $y = 2\cos x \sin 2x - \sin x \cos 2x$ gives:
$\frac{dy}{dx} = 2(-\sin x)\sin 2x + 2\cos x(2\cos 2x) - (\cos x\cos 2x + \sin x(-2\sin 2x))$
$\frac{dy}{dx} = -2\sin x\sin 2x + 4\cos x\cos 2x - \cos x\cos 2x + 2\sin x\sin 2x$
$\frac{dy}{dx} = 3\cos x\cos 2x$
Thus, $k = 3$.

2. Express the integral of $7\cos x\cos 2x$ using the result from part (a).

Since $\frac{d}{dx}(2\cos x \sin 2x - \sin x \cos 2x) = 3\cos x \cos 2x$,
$\int 3\cos x \cos 2x \, dx = 2\cos x \sin 2x - \sin x \cos 2x$
$\int 7\cos x \cos 2x \, dx = \frac{7}{3}(2\cos x \sin 2x - \sin x \cos 2x)$

3. Integrate the entire expression including $\sec^2 x$.

$\int \left(7\cos x\cos 2x + \sec^{2} x\right)\,dx = \frac{7}{3}(2\cos x \sin 2x - \sin x \cos 2x) + \tan x$

4. Evaluate the definite integral at the upper limit $\pi/4$.

At $x = \frac{\pi}{4}$:
$\frac{7}{3}\left(2\cos\frac{\pi}{4}\sin\frac{\pi}{2} - \sin\frac{\pi}{4}\cos\frac{\pi}{2}\right) + \tan\frac{\pi}{4}$
$= \frac{7}{3}\left(2\left(\frac{\sqrt{2}}{2}\right)(1) - \left(\frac{\sqrt{2}}{2}\right)(0)\right) + 1$
$= \frac{7}{3}(\sqrt{2}) + 1 = \frac{7\sqrt{2}}{3} + 1$

5. Evaluate the definite integral at the lower limit $0$ and find the final value.

At $x = 0$:
$\frac{7}{3}(2\cos 0\sin 0 - \sin 0\cos 0) + \tan 0 = 0$

Therefore,
$\int_{0}^{\pi/4}\left(7\cos x\cos 2x + \sec^{2} x\right)\,dx = \left(\frac{7\sqrt{2}}{3} + 1\right) - 0 = \frac{7\sqrt{2}}{3} + 1$ (shown)