If \[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos (\sin \,\,x)dx;{{I}_{2}}=\int\limits_{0}^{\frac{\pi }{2}}{\sin (\cos \,\,x)dx}}\] and \[{{I}_{3}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos x\,\,dx,}\] then

\[{{I}_{3}}>{{I}_{1}}>{{I}_{2}}\]

\[{{I}_{1}}>{{I}_{3}}>{{I}_{2}}\]

\[{{I}_{1}}>{{I}_{2}}>{{I}_{3}}\]

\[{{I}_{3}}>{{I}_{2}}>{{I}_{1}}\]

If \[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos (\sin \,\,x)dx;{{

1.	If \[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos (\sin \,\,x)dx;{{I}_{2}}=\int\limits_{0}^{\frac{\pi }{2}}{\sin (\cos \,\,x)dx}}\] and \[{{I}_{3}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos x\,\,dx,}\] then
A.	\[{{I}_{1}}>{{I}_{3}}>{{I}_{2}}\]
B.	\[{{I}_{3}}>{{I}_{1}}>{{I}_{2}}\]
C.	\[{{I}_{1}}>{{I}_{2}}>{{I}_{3}}\]
D.	\[{{I}_{3}}>{{I}_{2}}>{{I}_{1}}\]
Answer» B. \[{{I}_{3}}>{{I}_{1}}>{{I}_{2}}\]

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If \[{{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos (\sin \,\,x)dx;{{I}_{2}}=\int\limits_{0}^{\frac{\pi }{2}}{\sin (\cos \,\,x)dx}}\] and \[{{I}_{3}}=\int\limits_{0}^{\frac{\pi }{2}}{\cos x\,\,dx,}\] then

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