Let y=y(x) be the solution of the differential equation x tan( y x)dy=(y tan( y x)-x)dx, -1≤ x≤ 1,y( 1 2)= π 6. Then the area of the region bounded…

We have,$\frac{dy}{dx}=\frac{x(\frac{y}{x}\cdot \mathrm{tan}(\frac{y}{x})-1)}{x\mathrm{tan}(\frac{y}{x})}$dxdy​=xtan(xy​)x(xy​⋅tan(xy​)−1)​

$$\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\mathrm{cot}(\frac{y}{x})$$⇒dxdy​=xy​−cot(xy​)

Put $y=vx$y=vx

$$\Rightarrow \frac{dy}{dx}=v+x(\frac{dv}{dx})$$⇒dxdy​=v+x(dxdv​)

Now, we get

$$v+x(\frac{dv}{dx})=v-\mathrm{cot}v$$v+x(dxdv​)=v−cotv

$$\Rightarrow \int \mathrm{tan}vdv=-\int \frac{dx}{x}$$⇒∫tanvdv=−∫xdx​

$$\Rightarrow \mathrm{ln}|\mathrm{sec}v|=-\mathrm{ln}|x|+c$$⇒ln∣secv∣=−ln∣x∣+c

$$\Rightarrow \mathrm{ln}|\mathrm{sec}(\frac{y}{x})|=-\mathrm{ln}|x|+c$$⇒ln∣sec(xy​)∣=−ln∣x∣+c

$$\Rightarrow \mathrm{ln}|\mathrm{sec}(\frac{y}{x})|+\mathrm{ln}|x|=c$$⇒ln∣sec(xy​)∣+ln∣x∣=c

Now, $y(\frac{1}{2})=\frac{\pi }{6}$y(21​)=6π​, then

$$\mathrm{ln}|\mathrm{sec}(\frac{\pi }{3})|+\mathrm{ln}(\frac{1}{2})=c$$ln∣sec(3π​)∣+ln(21​)=c

$$\Rightarrow \mathrm{ln}2+\mathrm{ln}(\frac{1}{2})=c$$⇒ln2+ln(21​)=c

$$\Rightarrow \mathrm{ln}2-\mathrm{ln}2=c$$⇒ln2−ln2=c

$$\Rightarrow c=0$$⇒c=0

Hence,

$$\therefore \mathrm{sec}(\frac{y}{x})=\frac{1}{x}$$∴sec(xy​)=x1​

$$\Rightarrow \mathrm{cos}(\frac{y}{x})=x$$⇒cos(xy​)=x

$$\Rightarrow y=x{\mathrm{cos}}^{-1}(x)$$⇒y=xcos−1(x)

So, required bounded area

$$={\int }_{0}^{\frac{1}{\sqrt{2}}}(\underset{I}{\underset{⏟}{{\mathrm{cos}}^{-1}x}})\underset{\mathrm{II}}{x}dx$$=∫02​1​​(I⏟cos−1x​​)IIx​dx

$$={[{\mathrm{cos}}^{-1}x\cdot (\frac{{x}^{2}}{2})]}_{0}^{\frac{1}{\sqrt{2}}}+\frac{1}{2}{\int }_{0}^{\frac{1}{\sqrt{2}}}(\frac{{x}^{2}}{\sqrt{1-{x}^{2}}})dx$$=[cos−1x⋅(2x2​)]02​1​​+21​∫02​1​​(1−x2​x2​)dx

$$=\frac{\pi }{16}-\frac{1}{2}{\int }_{0}^{\frac{1}{\sqrt{2}}}(\frac{1-{x}^{2}-1}{\sqrt{1-{x}^{2}}})dx$$=16π​−21​∫02​1​​(1−x2​1−x2−1​)dx

$$=\frac{\pi }{16}-\frac{1}{2}{\int }_{0}^{\frac{1}{\sqrt{2}}}(\sqrt{1-{x}^{2}}-\frac{1}{\sqrt{1-{x}^{2}}})dx$$=16π​−21​∫02​1​​(1−x2​−1−x2​1​)dx

$$=\frac{\pi }{16}-\frac{1}{2}{[\frac{x}{2}\sqrt{1-{x}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}x-{\mathrm{sin}}^{-1}x]}_{0}^{\frac{1}{\sqrt{2}}}$$=16π​−21​[2x​1−x2​+21​sin−1x−sin−1x]02​1​​

$$=\frac{\pi }{16}-\frac{1}{2}{[\frac{x}{2}\sqrt{1-{x}^{2}}-\frac{1}{2}{\mathrm{sin}}^{-1}x]}_{0}^{\frac{1}{\sqrt{2}}}$$=16π​−21​[2x​1−x2​−21​sin−1x]02​1​​

$$=\frac{\pi }{16}-\frac{1}{2}[\frac{1}{4}-\frac{\pi }{8}]$$=16π​−21​[41​−8π​]

$=(\frac{\pi -1}{8})$=(8π−1​) sq. units