int cos (log x) d x=F(x)+C, where C is an arb | vedclass

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(C) Let $I = \int \cos (\log x) d x$. Substitute $\log x = t$,which implies $x = e^t$. Then,$dx = e^t dt$. Substituting these into the integral,we get

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

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(C) Let $I = \int \cos (\log x) d x$. Substitute $\log x = t$,which implies $x = e^t$. Then,$dx = e^t dt$. Substituting these into the integral,we get $I = \int e^t \cos t dt$. Using the standard integration formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} [a \cos(bx) + b \sin(bx)] + C$,where $a = 1$ and $b = 1$: $I = \frac{e^t}{1^2 + 1^2} [1 \cdot \cos t + 1 \cdot \sin t] + C$. $I = \frac{e^t}{2} [\cos t + \sin t] + C$. Substituting $t = \log x$ and $e^t = x$ back into the expression: $I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$. Thus,$F(x) = \frac{x}{2} [\cos(\log x) + \sin(\log x)]$.

$\int \cos (\log x) d x=F(x)+C,$ where $C$ is an arbitrary constant. Here,$F(x)$ is equal to

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