We have,
d y d x = x ( y x ⋅ tan ( y x ) − 1 ) x t a n ( y x ) \dfrac{dy}{dx} = \dfrac{x\left(\dfrac{y}{x} \cdot \tan \left(\dfrac{y}{x}\right) - 1\right)}{xtan\left(\dfrac{y}{x}\right)} d x d y = x t an ( x y ) x ( x y ⋅ tan ( x y ) − 1 ) ⇒ d y d x = y x − cot ( y x ) ⇒ \dfrac{dy}{dx} = \dfrac{y}{x} - \cot \left(\dfrac{y}{x}\right) ⇒ d x d y = x y − cot ( x y )
Put y = v x y = vx y = vx
⇒ d y d x = v + x ( d v d x ) ⇒ \dfrac{dy}{dx} = v + x\left(\dfrac{dv}{dx}\right) ⇒ d x d y = v + x ( d x d v )
Now, we get
v + x ( d v d x ) = v − cot v v + x\left(\dfrac{dv}{dx}\right) = v - \cot v v + x ( d x d v ) = v − cot v
⇒ ∫ t a n v d v = − ∫ d x x ⇒ ∫ tanvdv = - ∫ \dfrac{dx}{x} ⇒ ∫ t an v d v = − ∫ x d x
⇒ ln ∣ sec v ∣ = − ln ∣ x ∣ + c ⇒ \ln \left|\sec v\right| = - \ln \left|x\right| + c ⇒ ln ∣ sec v ∣ = − ln ∣ x ∣ + c
⇒ ln ∣ sec ( y x ) ∣ = − ln ∣ x ∣ + c ⇒ \ln \left|\sec\left(\dfrac{y}{x}\right)\right| = - \ln \left|x\right| + c ⇒ ln sec ( x y ) = − ln ∣ x ∣ + c
⇒ ln ∣ sec ( y x ) ∣ + ln ∣ x ∣ = c ⇒ \ln \left|\sec\left(\dfrac{y}{x}\right)\right| + \ln \left|x\right| = c ⇒ ln sec ( x y ) + ln ∣ x ∣ = c
Now, y ( 1 2 ) = π 6 y\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6} y ( 2 1 ) = 6 π
ln ∣ sec ( π 3 ) ∣ + ln ( 1 2 ) = c \ln \left|\sec\left(\dfrac{\pi}{3}\right)\right| + \ln \left(\dfrac{1}{2}\right) = c ln sec ( 3 π ) + ln ( 2 1 ) = c
⇒ ln 2 + ln ( 1 2 ) = c ⇒ \ln 2 + \ln \left(\dfrac{1}{2}\right) = c ⇒ ln 2 + ln ( 2 1 ) = c
⇒ ln 2 − ln 2 = c ⇒ \ln 2 - \ln 2 = c ⇒ ln 2 − ln 2 = c
⇒ c = 0 ⇒ c = 0 ⇒ c = 0
Hence,
∴ sec ( y x ) = 1 x ∴ \sec \left(\dfrac{y}{x}\right) = \dfrac{1}{x} ∴ sec ( x y ) = x 1
⇒ cos ( y x ) = x ⇒ \cos \left(\dfrac{y}{x}\right) = x ⇒ cos ( x y ) = x
⇒ y = x cos − 1 ( x ) ⇒ y = x\cos^{- 1}\left(x\right) ⇒ y = x cos − 1 ( x )
So, required bounded area
= ∫ 0 1 2 ( cos − 1 x ⏟ I ) x I I d x = {∫}_{0}^{\dfrac{1}{\sqrt{2}}}\left(\mathop{\mathop{\cos^{- 1}x}_{⏟}}_{I}\right)\mathop{x}_{II}dx = ∫ 0 2 1 ( cos − 1 x ⏟ I ) x II d x
= [ cos − 1 x ⋅ ( x 2 2 ) ] 0 1 2 + 1 2 ∫ 0 1 2 ( x 2 1 − x 2 ) d x = {\left[\cos^{- 1}x \cdot \left(\dfrac{{x}^{2}}{2}\right)\right]}_{0}^{\dfrac{1}{\sqrt{2}}} + \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{\sqrt{2}}}\left(\dfrac{{x}^{2}}{\sqrt{1 - {x}^{2}}}\right)dx = [ cos − 1 x ⋅ ( 2 x 2 ) ] 0 2 1 + 2 1 ∫ 0 2 1 ( 1 − x 2 x 2 ) d x
= π 16 − 1 2 ∫ 0 1 2 ( 1 − x 2 − 1 1 − x 2 ) d x = \dfrac{\pi}{16} - \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{\sqrt{2}}}\left(\dfrac{1 - {x}^{2} - 1}{\sqrt{1 - {x}^{2}}}\right)dx = 16 π − 2 1 ∫ 0 2 1 ( 1 − x 2 1 − x 2 − 1 ) d x
= π 16 − 1 2 ∫ 0 1 2 ( 1 − x 2 − 1 1 − x 2 ) d x = \dfrac{\pi}{16} - \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{\sqrt{2}}}\left(\sqrt{1 - {x}^{2}} - \dfrac{1}{\sqrt{1 - {x}^{2}}}\right)dx = 16 π − 2 1 ∫ 0 2 1 ( 1 − x 2 − 1 − x 2 1 ) d x
= π 16 − 1 2 [ x 2 1 − x 2 + 1 2 sin − 1 x − sin − 1 x ] 0 1 2 = \dfrac{\pi}{16} - \dfrac{1}{2}{\left[\dfrac{x}{2}\sqrt{1 - {x}^{2}} + \dfrac{1}{2}\sin^{- 1}x - \sin^{- 1}x\right]}_{0}^{\dfrac{1}{\sqrt{2}}} = 16 π − 2 1 [ 2 x 1 − x 2 + 2 1 sin − 1 x − sin − 1 x ] 0 2 1
= π 16 − 1 2 [ x 2 1 − x 2 − 1 2 sin − 1 x ] 0 1 2 = \dfrac{\pi}{16} - \dfrac{1}{2}{\left[\dfrac{x}{2}\sqrt{1 - {x}^{2}} - \dfrac{1}{2}\sin^{- 1}x\right]}_{0}^{\dfrac{1}{\sqrt{2}}} = 16 π − 2 1 [ 2 x 1 − x 2 − 2 1 sin − 1 x ] 0 2 1
= π 16 − 1 2 [ 1 4 − π 8 ] = \dfrac{\pi}{16} - \dfrac{1}{2}\left[\dfrac{1}{4} - \dfrac{\pi}{8}\right] = 16 π − 2 1 [ 4 1 − 8 π ]
= ( π − 1 8 ) = \left(\dfrac{\pi - 1}{8}\right) = ( 8 π − 1 )
