Let y = y x be the solution of the differential equation x   tan y x d y = y   tan y x - x d x , -…

$\frac{dy}{dx}=\frac{x\left({\displaystyle \frac{y}{x}}·\tan \left({\displaystyle \frac{y}{x}}\right)-1\right)}{x\tan \left({\displaystyle \frac{y}{x}}\right)}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\cot \left(\frac{y}{x}\right)$

Put $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\left(\frac{dv}{dx}\right)$

Now, we get

$v+x\left(\frac{dv}{dx}\right)=v-\cot v$

$\Rightarrow \int \tan vdv=-\int \frac{dx}{x}$

$\Rightarrow \ln \left|\sec v\right|=-\ln \left|x\right|+c$

$\Rightarrow \ln \left|\sec \left(\frac{y}{x}\right)\right|=-\ln \left|x\right|+c$

$\Rightarrow \ln \left|\sec \left(\frac{y}{x}\right)\right|+\ln \left|x\right|=c$

Now, $y\left(\frac{1}{2}\right)=\frac{\pi }{6}$, then

$\ln \left|\sec \left(\frac{\pi }{3}\right)\right|+\ln \left(\frac{1}{2}\right)=c$

$\Rightarrow \ln 2+\ln \left(\frac{1}{2}\right)=c$

$\Rightarrow \ln 2-\ln 2=c$

$\Rightarrow c=0$

Hence,

$\therefore \sec \left(\frac{y}{x}\right)=\frac{1}{x}$

$\Rightarrow \cos \left(\frac{y}{x}\right)=x$

$\Rightarrow y=x{\cos }^{-1}\left(x\right)$

So, required bounded area

$={\int }_0^{\frac{1}{\sqrt{2}}}\left(\underset{\mathrm{I}}{\underbrace{{\cos }^{-1}x}}\right)\underset{\mathrm{II}}{x}dx$

$={\left[{\cos }^{-1}x·\left(\frac{x^2}{2}\right)\right]}_0^{\frac{1}{\sqrt{2}}}+\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{x^2}{\sqrt{1-x^2}}\right)dx$

$=\frac{\pi }{16}-\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{1-x^2-1}{\sqrt{1-x^2}}\right)dx$

$=\frac{\pi }{16}-\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{2}}}\left(\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}}\right)dx$

$=\frac{\pi }{16}-\frac{1}{2}{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-{\sin }^{-1}x\right]}_0^{\frac{1}{\sqrt{2}}}$

$=\frac{\pi }{16}-\frac{1}{2}{\left[\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{\sin }^{-1}x\right]}_0^{\frac{1}{\sqrt{2}}}$

$=\frac{\pi }{16}-\frac{1}{2}\left[\frac{1}{4}-\frac{\pi }{8}\right]$

$=\left(\frac{\pi -1}{8}\right)$ sq. units