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(C) Let $I = \int \sin(\log x) \, dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t \, dt$. Then,$I = \int e^t \sin t \, dt$. Using i

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$\frac{1}{2}x[\sin(\log x) - \cos(\log x)] + C$

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(C) Let $I = \int \sin(\log x) \, dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t \, dt$. Then,$I = \int e^t \sin t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,let $u = \sin t$ and $dv = e^t \, dt$. Then $du = \cos t \, dt$ and $v = e^t$. $I = e^t \sin t - \int e^t \cos t \, dt$. Applying integration by parts again to $\int e^t \cos t \, dt$: $I = e^t \sin t - [e^t \cos t - \int e^t (-\sin t) \, dt]$. $I = e^t \sin t - e^t \cos t - I$. $2I = e^t(\sin t - \cos t)$. $I = \frac{1}{2} e^t(\sin t - \cos t) + C$. Substituting $t = \log x$ and $e^t = x$: $I = \frac{1}{2} x[\sin(\log x) - \cos(\log x)] + C$.

$\int \sin(\log x) \, dx = $

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