A) \[\sin \,(\log 3)\]
B) \[\sin \,(\log 2)\]
C) \[\cos \,(\log 3)\]
D) None of these
Correct Answer: B
Solution :
Put \[t=\log x\Rightarrow dt=\frac{1}{x}dx\]. As \[x=2\Rightarrow t=\log 2\]and \[x=1\Rightarrow t=0\], we have \[\int_{1}^{2}{\frac{\cos (\log x)}{x}}dx=-\int_{0}^{\log 2}{\cos t\,dt}=[\sin t]_{0}^{\log 2}\]\[=\sin (\log 2)\].
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