Evaluate: ∫ [sin(log x) + cos(log x)] dx

Put $\log x = t$, so $x = e^t$ and $dx = e^t dt$. $I = \int e^t(\sin t + \cos t)\, dt$ Using $\int e^t[f(t) + f'(t)]\, dt = e^t f(t)$ with $f(t) = \sin t$: $I = e^t \sin t + C$ Hence $\int [\sin(\log x) + \cos(\log x)]\, dx = x\sin(\log x) + C$.