int_{1}^{2} frac{cos(log x)}{x} , dx = | vedclass

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(B) Let $t = \log x$. Then,$dt = \frac{1}{x} dx$. When $x = 1$,$t = \log 1 = 0$. When $x = 2$,$t = \log 2$. Substituting these into the integral,we ge

Vedclass · Accepted Answer

$\sin(\log 2)$

Vedclass · Answer

(B) Let $t = \log x$. Then,$dt = \frac{1}{x} dx$. When $x = 1$,$t = \log 1 = 0$. When $x = 2$,$t = \log 2$. Substituting these into the integral,we get: $\int_{1}^{2} \frac{\cos(\log x)}{x} \, dx = \int_{0}^{\log 2} \cos(t) \, dt$. Evaluating the integral: $\int_{0}^{\log 2} \cos(t) \, dt = [\sin(t)]_{0}^{\log 2} = \sin(\log 2) - \sin(0) = \sin(\log 2) - 0 = \sin(\log 2)$. Thus,the correct option is $B$.

$\int_{1}^{2} \frac{\cos(\log x)}{x} \, dx = $

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