Explanation

(a)
Let $I = \int_{}^{}{\cos{(\log_{e}{x)dx = \int_{}^{}{\cos{(\log_{e}{x) \bullet 1dx}}}}}}$
$= \cos{(\log_{e}{x)x - \int_{}^{}{\frac{{- sin}{\log_{e}{x)}}}{x} \bullet x\ dx}}}$
$= x\cos{(\log_{e}{x) + \int_{}^{}{\sin{(\log_{e}{x)dx}}}}}$
$= x\cos{(\log_{e}{x) + \int_{}^{}{\sin{(\log_{e}{x) \bullet 1\ dx}}}}}$
$= x\cos{(\log_{e}{x) - \sin{(\log_{e}{x)x}} - \int_{}^{}{\frac{\cos{(\log_{e}{x)}}}{x}x\ dx}}}$
$= x\cos{(\log_{e}{x)}} + x\sin{(\log_{e}{x) - I}}\$
$\Rightarrow 2I = x\lbrack\cos{(\log_{e}{x)}} + \sin{(\log_{e}{x)\rbrack}}$
$\Rightarrow I = \frac{x}{2}\lbrack\cos{(\log_{e}{x)}} + \sin{(\log_{e}{x)\rbrack}}$