Question

Evaluate : ${\int }_1^3\left[\frac{\cos \left(\log x\right)}{x}\right]dx$

• A. $1$
• B. $\cos \left(\log 3\right)$
• C. $\sin \left(\log 3\right)$
• D. $\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is C

$\sin \left(\log 3\right)$

Evaluating the given integral

$I={\int }_1^3\left[\frac{\cos \left(\log x\right)}{x}\right]dx$

Let $\log x=t$

Differentiating with respect to $x$

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

if $x=1$ then $t=0$

if $x=3$ then $t=\log 3$⁠⁠⁠⁠⁠⁠

Thus, We have :

$I={\int }_0^{\log 3}\cos tdt$

$={\left[\sin t\right]}_0^{\log 3}$

$=\sin \left(\log 3\right)-\sin \left(0\right)$

$I=\sin \left(\log 3\right)$

Hence, Option C is the correct answer