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

$\begin{array}{l}slope=\frac{\mathrm{dy}}{\mathrm{dx}}=2\tan \mathrm{xcos}\mathrm{x}-2\tan \mathrm{x}\cdot \mathrm{y}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+(2\tan \mathrm{x})\mathrm{y}=2\sin \mathrm{x}\end{array}$

Integrating factor $={\mathrm{e}}^{\int 2\tan \mathrm{xdx}}=\frac{1}{{\cos }^2\mathrm{x}}$

$\begin{array}{l}\mathrm{y}\left(\frac{1}{{\cos }^2\mathrm{x}}\right)=\int \frac{2\sin \mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{dx}\\ {\mathrm{ysec}}^2\mathrm{x}=2 secx+\mathrm{C}\end{array}$

$\mathrm{y}=2\cos \mathrm{x}+{\mathrm{Ccos}}^2\mathrm{x}$

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

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

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

$\begin{array}{l}{\int }_0^{\mathrm{\pi }/2} \mathrm{ydx}=2{\int }_0^{\mathrm{\pi }/2} \cos \mathrm{xdx}-2\sqrt{2}{\int }_0^{\mathrm{\pi }/2} {\cos }^2\mathrm{xdx}\\ =[2\sin \mathrm{x}{]}_0^{\mathrm{\pi }/2}-2\sqrt{2}{\left[\frac{\mathrm{x}}{2}+\frac{\sin 2\mathrm{x}}{4}\right]}_0^{\mathrm{\pi }/2}\\ =2-\frac{\mathrm{\pi }}{\sqrt{2}}\end{array}$