A) \[si{{n}^{-1}}y=si{{n}^{-1}}x+c\]
B) \[2si{{n}^{-1}}y=\sqrt{1-{{x}^{2}}}+si{{n}^{-1}}x+c\]
C) \[2si{{n}^{-1}}y=x\sqrt{1-{{x}^{2}}}+si{{n}^{-1}}x+c\]
D) \[2si{{n}^{-1}}y=x\sqrt{1-{{x}^{2}}}+{{\cos }^{-1}}x+c\]
Correct Answer: D
Solution :
[c] \[\because \,\,\,\frac{dy}{dx}=\sqrt{1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}}\] \[\frac{dy}{dx}=\sqrt{(1-{{x}^{2}})(1-{{y}^{2}});}\Rightarrow \frac{dy}{\sqrt{1-{{y}^{2}}}}=\sqrt{1-{{x}^{2}}}.dx\] \[\Rightarrow \int{\frac{dy}{\sqrt{1-{{y}^{2}}}}=\int{\sqrt{1-{{x}^{2}}}.dx}}\] [integrating b/s] \[\Rightarrow {{\sin }^{-1}}\left( \frac{y}{1} \right)=\frac{x}{2}\sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}\left( \frac{x}{1} \right)+c\] \[\Rightarrow 2{{\sin }^{-1}}y=x\sqrt{1-{{x}^{2}}}+{{\sin }^{-1}}x+c\] where c is an arbitrary constant.
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