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(A) Let $I = \int \cos(\ln x) \, dx$.Using integration by parts, let $u = \cos(\ln x)$ and $dv = dx$. Then $du = -\sin(\ln x) \cdot \frac{1}{x} \, dx$

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$\frac{x}{2} \{\cos(\ln x) + \sin(\ln x)\} + c$

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(A) Let $I = \int \cos(\ln x) \, dx$.Using integration by parts, let $u = \cos(\ln x)$ and $dv = dx$. Then $du = -\sin(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.$I = x \cos(\ln x) - \int x \cdot (-\sin(\ln x)) \cdot \frac{1}{x} \, dx$$I = x \cos(\ln x) + \int \sin(\ln x) \, dx$.Now, integrate $\int \sin(\ln x) \, dx$ by parts again, with $u = \sin(\ln x)$ and $dv = dx$. Then $du = \cos(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.$\int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \cdot \cos(\ln x) \cdot \frac{1}{x} \, dx = x \sin(\ln x) - I$.Substituting this back into the equation for $I$:$I = x \cos(\ln x) + x \sin(\ln x) - I$$2I = x \{\cos(\ln x) + \sin(\ln x)\}$$I = \frac{x}{2} \{\cos(\ln x) + \sin(\ln x)\} + c$.

$I = \int \cos(\ln x) \, dx$. Then $I =$

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