The integral $ \int {\cos ({{\log }_e}x)dx} $ is equal to:
(Where C is a constant of integration)
(1) $ \dfrac{x}{2}[\sin ({\log _e}x) - \cos ({\log _e}x)] + C $
(2) $ \dfrac{x}{2}[\cos ({\log _e}x) + \sin ({\log _e}x)] + C $
(3) $ x[\cos ({\log _e}x) + \sin ({\log _e}x)] + C $
(4) $ x[\cos ({\log _e}x) - \sin ({\log _e}x)] + C $

Hint: We know that for integral problems angle of trigonometric functions must be linear or simple but if not then we use substitution method to convert angle of trigonometric function into simpler form and then using by part method of integration on the result so obtained to find required solution of the problem.
Formulas used: By part formula: $ 1st\int {2nd\,\,dx - \int {\left[ {\dfrac{d}{{dx}}(1st).\int {2nd} \,\,dx} \right]dx} } $ where $ 1st\,\,and\,\,2nd $ function can be calculated by using ILATE where I stands for inverse, L for logarithm, A for algebra, T for trigonometry and E for exponential functions. Which letter comes before the other in ILATE will be taken as first and the other will be taken as second function.

Complete step by step solution:
Let the given integral be equal to I.
Therefore we have
 $ I = \int {\cos ({{\log }_e}x)dx} $
Let $ {\log _e}x = t $
Taking antilog both sides. We have,
 $ x = {e^t} $
Now, differentiating above with respect to x to obtain the value of dx.
 $ dx = {e^t}dt $
Using values so obtained in I. We have,
 $ I = \int {\cos (t).{e^t}dt} $
 $ I = \int {{e^t}.\cos t\,\,dt} $
Now, applying the method of by part on integral. We have,
 $ I = \cos t\int {{e^t}dt - \int {\left\{ {\dfrac{d}{{dt}}(\cos t)\int {{e^t}dt} } \right\}dt} } $
\[
\Rightarrow I = \cos t({e^t}) - \int {( - \sin t){e^t}} dt \\
\Rightarrow I = \cos t({e^t}) + \int {\sin t.{e^t}dt} \\
 \]
Again applying by part method on the second part of the above equation. We have,
 $ I = \cos t({e^t}) + \left[ {\sin t\int {{e^t}dt} - \int {\left\{ {\dfrac{d}{{dt}}(\sin t)\int {{e^t}} dt} \right\}dt} } \right] $
\[
\Rightarrow I = \cos t({e^t}) + \sin t({e^t}) - \int {\cos t.{e^t}dt} \;
  or
\Rightarrow I = \cos t({e^t}) + \sin t({e^t}) - I \\
\Rightarrow I + I = {e^t}\left( {\cos t + \sin t} \right) \\
\Rightarrow 2I = {e^t}\left( {\cos t + \sin t} \right) \\
\Rightarrow I = \dfrac{1}{2}{e^t}\left( {\cos t + \sin t} \right) + C \\
 \]
Now, using value of t in term of x in above equation we have,
 $ I = \dfrac{1}{2}(x)\left\{ {\sin ({{\log }_e}x) + \cos ({{\log }_e}x)} \right\} $ + C
Therefore, required value of integral $ \int {\cos ({{\log }_e}x)dx} $ is $ \dfrac{1}{2}(x)\left\{ {\sin ({{\log }_e}x) + \cos ({{\log }_e}x)} \right\} $
Hence, from given options we see that the correct option is option $ \left( 2 \right) $ .
So, the correct answer is “Option 2”.

Note: In integral having two functions in a product where one is exponential and other is trigonometric function. To find integration of this type of problems we used the ‘I’ method using part method of integration. In this we first give the given integral ‘I’ and then apply part of the problem until we obtain ‘I’ on the right hand side. Then shifting ‘I’ obtained on right hand side to left hand side to obtain respective value of ‘I’ or solution of required integral.