Question

The value of${\int }_1^2\frac{\cos \left(\log x\right)}{x}dx$ is equal to

• A. $2\sin \left(\log 2\right)$
• B. $\sin \left(\log 2\right)$
• C. $\sin \log \left(\frac{1}{2}\right)$
• D. None of these

Answer 1

The correct option is B $\sin \left(\log 2\right)$

${\int }_1^2\frac{\cos \left(\log x\right)}{x}dx$

Substitute $\log x=t$

$\frac{1}{x}dx=dt$

$I={\int }_{\log 1}^{\log 2}\cos tdt$

$I={\left[\sin t\right]}_{\log 1}^{\log 2}$

$I=\sin (\log 2)$