Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $- 1 \le x \le 1$, $y\left( {{1 \over 2}} \right) = {\pi \over 6}$. Then the area of the region bounded by the curves x = 0, $x = {1 \over {\sqrt 2 }}$ and y = y(x) in the upper half plane is :

We have,<br><br>$${{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}$$<br><br>$\therefore$ ${{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)$<br><br>Put ${y \over x} = v$<br><br>$\Rightarrow y = vn$<br><br>$\therefore$ ${{dy} \over {dx}} = v + {{ndv} \over {dx}}$<br><br>Now, we get<br><br>$v + n{{dv} \over {dx}} = v - \cot (v)$<br><br>$\Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}}$<br><br>$\therefore$ $$\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c$$<br><br>As, $\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0$<br><br>$\therefore$ $\sec \left( {{y \over x}} \right) = {1 \over x}$<br><br>$\Rightarrow \cos \left( {{y \over x}} \right) = x$<br><br>$\therefore$ $y = x{\cos ^{ - 1}}(x)$<br><br>So, required bounded area<br><br>$$ = \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)} $$ (I. B. P.)<br><br>$\therefore$ Option (1) is correct.