Integrating $\int(\sin(\log x) + \cos(\log x)) dx$

Question

Integrating $$\int(\sin(\log x) + \cos(\log x)) dx$$

I did the sum till here:

I'm not getting what to do next. I tried LIATE method, but it's not working.

Please help me out

Answer 1

Use the identity integral of

$$\int e^x( f(x) +f'(x))= f(x)$$

which should give you

$x\sin(\log x) + C$

Answer 2

Let \begin{equation*} I = \int {\big(\sin{(\log{x})} + \cos{(\log{x})}\big)\,\mathrm{d}x}. \end{equation*} Following your approach, \begin{equation*} I = \int {e^{t}(\sin{t} + \cos{t})\, \mathrm{d}t} \end{equation*} where $t = \log{x}$.

Now applying the integration by parts formula $\int u \mathrm\,{d}v = uv - \int v\,\mathrm{d}u$ with $u = e^{t}$ and $\mathrm{d}v = (\sin{t} + \cos{t})\,\mathrm{d}t$, we have \begin{equation*} I = e^{t}(\sin{t} - \cos{t}) + \int e^{t}(\cos{t} - \sin{t})\,\mathrm{d}t. \tag*{(1)} \end{equation*} We can again use integration by parts on the second term of the right-hand side to get \begin{equation*} \int e^{t}(\cos{t} - \sin{t})\,\mathrm{d}t= e^{t}(\sin{t} + \cos{t})-\int e^{t}(\sin{t} + \cos{t})\,\mathrm{d}t = e^{t}(\sin{t} + \cos{t}) - I. \end{equation*} Notice we have regenerated our starting integral, $I$, which happens frequently when finding indefinite integrals involving both exponential and trigonometric functions. Placing the above equality into $(1)$ gives \begin{equation*} 2I = 2e^{t}\sin{t} + C \end{equation*} and so \begin{equation*} I = x\sin{(\log{x})} + C. \end{equation*}