Let the slope of the tangent to a curve y = f(x) at (x, y) be given...

Given,

$\dfrac{dy}{dx} = 2\tan x \cos x - 2\tan x \cdot y$

$\dfrac{dy}{dx} + \left(2\tan x\right)y = 2\sin x$

Now solving linear differential equation by finding Integrating factor $= {e}^{∫ 2tanxdx} = \dfrac{1}{\cos^{2}x}$

Solution is given by,

$y\left(\dfrac{1}{\cos^{2}x}\right) = ∫ \dfrac{2\sin x}{\cos^{2}x}dx$

$⇒ y{sec}^{2}x = \dfrac{2}{\cos x} + C$

$⇒ y = 2\cos x + C\cos^{2}x$

Passes through \left(\dfrac{\pi}{4}, 0\right)$</p><p>$⇒ 0 = \sqrt{2} + \dfrac{C}{2} ⇒ C = - 2\sqrt{2}$</p><p>So,$f\left(x\right) = 2\cos x - 2\sqrt{2}\cos^{2}x$: Required curve</p><p>Now,${∫}_{0}^{\dfrac{\pi}{2}}ydx = 2{∫}_{0}^{\dfrac{\pi}{2}}cosxdx - 2\sqrt{2}{∫}_{0}^{\dfrac{\pi}{2}}\cos^{2}xdx$</p><p>$= {\left[2\sin x\right]}_{0}^{\dfrac{\pi}{2}} - 2\sqrt{2}{\left[\dfrac{x}{2} + \dfrac{\sin 2x}{4}\right]}_{0}^{\dfrac{\pi}{2}}$</p><p>$= 2 - \dfrac{\pi}{\sqrt{2}}$