Let the slope of the tangent to a curve  $y=f\left(x\right)$ at  $\left(x,y\right)$  be given by  $2Tanx\left(\cos x-y\right)$. If the curve passes through $\left(\frac{\pi }{4},0\right)$  then  $\underset{0}{\overset{\pi /2}{\int }}ydx=$ 

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

$=e^{{\int }_{ }^{ }2Tanxdx}={\sec }^{2}x$

$y\left({\sec }^{2}x\right)={\int }_{ }^{ }{\sec }^{2}x\left(2\sin x\right)dx$

$y{\sec }^{2}x=2\sec x+c$

$\left(\frac{\pi }{4},0\right)\Rightarrow 0=2\sqrt{2}+c\Rightarrow c=-2\sqrt{2}$

$\left(\frac{\pi }{4},0\right)\Rightarrow 0=2\sqrt{2}+c\Rightarrow c=-2\sqrt{2}$

$y{\sec }^{2}x=2\sec x-2\sqrt{2}$

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

$\underset{0}{\overset{\pi /2}{\int }}ydx=2-\frac{\pi }{\sqrt{2}}$