Using a suitable substitution, find
$$\int \sqrt { 1 - x ^ { 2 } } d x$$
Show that the differential equation
$$\frac { d y } { d x } = 2 \sqrt { 1 - x ^ { 2 } - y ^ { 2 } + x ^ { 2 } y ^ { 2 } }$$
given that \(y = 0\) when \(x = 0 , | x | < 1\) and \(| y | < 1\), has the solution
$$y = x \cos \left( x \sqrt { 1 - x ^ { 2 } } \right) + \sqrt { 1 - x ^ { 2 } } \sin \left( x \sqrt { 1 - x ^ { 2 } } \right) .$$
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