$\int \cos (\log x) d x=F(x)+C,$ where $C$ is an arbitrary constant. Here, $F(x)$ is equal to
$\int \cos (\log x) d x=F(x)+C,$ where $C$ is an arbitrary constant. Here, $F(x)$ is equal to
$x[\cos (\log x)+\sin (\log x)]$
$x[\cos (\log x)-\sin (\log x)]$
$\frac{x}{2}[\cos (\log x)+\sin (\log x)]$
$\frac{x}{2}[\cos (\log x)-\sin (\log x)]$
Solution
Let $I=\int \cos (\log x) d x$
Put $\log x=t$
$\Rightarrow \quad x=e^{t}$
$\begin{aligned} \therefore \quad d x &=e^{t} d t \\ \therefore \quad I &=\int e^{t} \cos t d t \\ &=\frac{e^{t}}{1^{2}+1^{2}}[\cos t+\sin t]+C \end{aligned}$
$\left[\because \int e^{a x} \cos b x d x=\frac{e^{a x}}{a^{2}+b^{2}}[a \cos b x+b \sin b x]+c\right]$
$\Rightarrow \quad I=\frac{e^{t}}{2}[\cos t+\sin t]+C$
$\quad \quad=\frac{x}{2}[\cos (\log x)+\sin (\log x]+C$
$\therefore f(x)=\frac{x}{2}[\cos (\log x)+\sin (\log x)]$