The integral $$\int \cos(\ln x) dx$$, is equal to

We wish to evaluate the indefinite integral

$$\int \cos(\ln x)\,dx.$$

Because the argument of the cosine is $$\ln x$$, it is natural to remove the logarithm by the substitution

$$t=\ln x.$$

Then we have $$x=e^{t}$$ and, differentiating both sides,

$$dx=\frac{dx}{dt}\,dt=e^{t}\,dt=x\,dt.$$

Substituting these expressions into the integral gives

$$\int \cos(\ln x)\,dx=\int \cos(t)\,(e^{t}\,dt)=\int e^{t}\cos t\,dt.$$

Now we have to integrate the product $$e^{t}\cos t$$. For such integrals the standard method is integration by parts applied twice, but an even quicker way is to recall the known formula

$$\int e^{at}\cos(bt)\,dt=\frac{e^{at}}{a^{2}+b^{2}}\left(a\cos bt+b\sin bt\right)+C.$$

In our case $$a=1$$ and $$b=1$$, so

$$\int e^{t}\cos t\,dt=\frac{e^{t}}{1^{2}+1^{2}}\Bigl(1\cdot\cos t+1\cdot\sin t\Bigr)+C =\frac{e^{t}}{2}\bigl(\cos t+\sin t\bigr)+C.$$

Now we back-substitute $$t=\ln x$$ and $$e^{t}=x$$:

$$\int \cos(\ln x)\,dx=\frac{x}{2}\left(\cos(\ln x)+\sin(\ln x)\right)+C.$$

Hence, the correct answer is Option D.