A) \[{{\sin }^{2}}x\]
B) \[\frac{2}{\sin x}\]
C) \[\log \left| \sin \,\,x \right|\]
D) \[\frac{1}{{{\sin }^{2}}x}\]
Correct Answer: A
Solution :
[a] Given differential equation is \[\sin x\frac{dy}{dx}+2y\cos x=1;\Rightarrow \frac{dy}{dx}+2y\frac{\cos x}{\sin x}=\frac{1}{\sin x}\] \[\Rightarrow \frac{dy}{dx}+(2cotx)y=cosecx\] I.F. \[={{e}^{\int{2\cot xdx}}}={{e}^{\int{2\left( \frac{\cos x}{\sin x} \right)dx}}}={{e}^{2\log \sin x}}\] \[={{e}^{{{\log }_{e}}{{(sinx)}^{2}}}}={{(sinx)}^{2}}\]
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