Let the slope of the tangent to a curve $y=f\left(x\right)$ at $\left(x,y\right)$ be given by $2\tan x\left(\cos x-y\right)$. if the curve passes through the point $\left(\frac{\pi }{4},0\right)$, then the value of ${\int }_0^{\frac{\pi }{2}}ydx$ is equal to

Given,

$\frac{dy}{dx}=2\tan x\cos x-2\tan x·y$

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

Now solving linear differential equation by finding Integrating factor $={\mathrm{e}}^{\int 2\tan xdx}=\frac{1}{{\cos }^{2}x}$

Solution is given by,

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

$\Rightarrow y{\sec }^{2}x=\frac{2}{\cos x}+C$

$\Rightarrow y=2\cos x+C{\cos }^{2}x$

Passes through $\left(\frac{\pi }{4},0\right)$

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

So, $f\left(x\right)=2\cos x-2\sqrt{2}{\cos }^{2}x$: Required curve

Now, ${\int }_0^{\frac{\pi }{2}}ydx=2{\int }_0^{\frac{\pi }{2}}\cos xdx-2\sqrt{2}{\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

$={\left[2\sin x\right]}_0^{\frac{\pi }{2}}-2\sqrt{2}{\left[\frac{x}{2}+\frac{\sin 2x}{4}\right]}_0^{\frac{\pi }{2}}$

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