A) \[\frac{1}{2}x\{\cos ({{\log }_{e}}x)+\sin ({{\log }_{e}}x)\}\]
B) \[x\{\cos ({{\log }_{e}}x)+\sin ({{\log }_{e}}x)\}\]
C) \[\frac{1}{2}x\{\cos ({{\log }_{e}}x)-\sin ({{\log }_{e}}x)\}\]
D) \[x\{\cos ({{\log }_{e}}x)-\sin ({{\log }_{e}}x)\}\]
Correct Answer: A
Solution :
Let \[I=\int{\cos ({{\log }_{e}}x)\,dx}\]\[=\int{\cos ({{\log }_{e}}x)\,.\,1\,dx}\] \[I=\cos ({{\log }_{e}}x).\,x-\int{\frac{-\sin ({{\log }_{e}}x)}{x}}.\,x\,\,dx\] \[=x\cos ({{\log }_{e}}x)+\int{\sin ({{\log }_{e}}x)}\,\,dx\] \[=x\cos \,({{\log }_{e}}x)+\int{\sin \,({{\log }_{e}}x)}\,\,1\,\,dx\] \[=x\cos ({{\log }_{e}}x)+\sin ({{\log }_{e}}x).\,x-\int{\frac{\cos ({{\log }_{e}}x)}{x}x\,dx}\] \[=x\,\cos ({{\log }_{e}}x)+x\sin ({{\log }_{e}}x)-I\] \[\Rightarrow 2I=x\,[\cos \,({{\log }_{e}}x)+\sin \,({{\log }_{e}}x)]\] \[\Rightarrow I=\frac{x}{2}\,[\cos \,\,({{\log }_{e}}x)+\sin \,({{\log }_{e}}x)]\].
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