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

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$\frac{x}{2}(\cos (\log x)+\sin (\log x))+c$,(where $c$ is a constant of integration)

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(C) Let $I = \int \cos (\log x) d x$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \cos t dt$. Using the integration by parts formula $\int e^t f(t) dt = e^t \int f(t) dt - \int (e^t \int f(t) dt) dt$ or the standard result $\int e^t (\cos t + \sin t) dt = e^t \cos t + C$,we apply integration by parts: $I = \cos t \cdot e^t - \int (-\sin t) \cdot e^t dt = e^t \cos t + \int e^t \sin t dt$. Applying integration by parts again to $\int e^t \sin t dt$: $I = e^t \cos t + [\sin t \cdot e^t - \int \cos t \cdot e^t dt] = e^t \cos t + e^t \sin t - I$. Thus,$2I = e^t (\cos t + \sin t) + c_1$. $I = \frac{e^t}{2} (\cos t + \sin t) + c$. Substituting back $t = \log x$ and $e^t = x$: $I = \frac{x}{2} (\cos (\log x) + \sin (\log x)) + c$.

$\int \cos (\log x) d x=$

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