Let y=y(x) solve x!(yx)dy = [y!(yx)-x]dx, -1 x1, y!(12)=6.

Question

Let $y=y(x)$ solve $x	an\!\left(\dfrac{y}{x}ight)dy = \left[y	an\!\left(\dfrac{y}{x}ight)-xight]dx$, $-1\le x\le1$, $y\!\left(\dfrac12ight)=\dfrac{\pi}{6}$. The area of the region bounded by $x=0$, $x=\dfrac{1}{\sqrt2}$, and $y=y(x)$ in the upper half plane is:

Accepted Answer

With $y=vx$: $\dfrac{dv}{-\cot v}=\dfrac{dx}{x}$... integrating gives $\sec(y/x)=1/x$, i.e. $\cos(y/x)=x$, so $y=x\cos^{-1}x$.

$$\text{Area}=\int_0^{1/\sqrt2}x\cos^{-1}x\,dx = \frac{\pi-1}{8}.$$

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