Find out the area bounded by the curve y = _ 1/8 ^ ^2 x ( ^ -1 t )

MATHEMATICS · Subjective · CBSE · Mathematics · Differential Equations · Differential Equations

Problem

Find out the area bounded by the curve y=∫1/8sin⁡2x(sin⁡−1t) dt+∫1/8cos⁡2x(cos⁡−1t) dt(0≤x≤π/2)y = \int_{1/8}^{\sin^2 x} (\sin^{-1} \sqrt{t})\, dt + \int_{1/8}^{\cos^2 x} (\cos^{-1} \sqrt{t})\, dt \quad (0 \le x \le \pi/2) and the curve satisfying the differential equation y(x+y3) dx=x(y3−x) dyy(x + y^3)\, dx = x(y^3 - x)\, dy passing through (4,−2)(4, -2).

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