Let the slope of the tangent to a curve $$y = f(x)$$ at $$(x, y)$$ be given by $$2 \tan x(\cos x - y)$$. If the curve passes through the point $$\left(\frac{\pi}{4}, 0\right)$$, then the value of $$\int_0^{\pi/2} y \, dx$$ is equal to

The slope of the tangent is given by $$\frac{dy}{dx} = 2\tan x(\cos x - y)$$, and the curve passes through $$(\pi/4, 0)$$.

Write the ODE in standard linear form:

$$\frac{dy}{dx} + 2\tan x \cdot y = 2\tan x \cos x = 2\sin x$$

Find the integrating factor:

$$\mu = e^{\int 2\tan x\,dx} = e^{-2\ln|\cos x|} = \frac{1}{\cos^2 x} = \sec^2 x$$

Multiply through by the integrating factor:

$$\frac{d}{dx}[y\sec^2 x] = 2\sin x \sec^2 x = \frac{2\sin x}{\cos^2 x}$$

Integrate:

$$y\sec^2 x = \int \frac{2\sin x}{\cos^2 x}\,dx = \frac{2}{\cos x} + C = 2\sec x + C$$

$$y = 2\cos x + C\cos^2 x$$

Apply initial condition $$y(\pi/4) = 0$$:

$$0 = 2\cos(\pi/4) + C\cos^2(\pi/4) = 2 \cdot \frac{1}{\sqrt{2}} + C \cdot \frac{1}{2}$$

$$0 = \sqrt{2} + \frac{C}{2}$$

$$C = -2\sqrt{2}$$

So $$y = 2\cos x - 2\sqrt{2}\cos^2 x$$.

Compute $$\int_0^{\pi/2} y\,dx$$:

$$\int_0^{\pi/2} y\,dx = \int_0^{\pi/2} (2\cos x - 2\sqrt{2}\cos^2 x)\,dx$$

$$= 2[\sin x]_0^{\pi/2} - 2\sqrt{2}\int_0^{\pi/2}\cos^2 x\,dx$$

$$= 2(1 - 0) - 2\sqrt{2} \cdot \frac{\pi}{4}$$

$$= 2 - \frac{\pi\sqrt{2}}{2} = 2 - \frac{\pi}{\sqrt{2}}$$

The correct answer is Option B: $$\boxed{2 - \dfrac{\pi}{\sqrt{2}}}$$.