Integral of sin(log(x))/ dx

1. Let $u = \log{\left(x \right)}$.

   Then let $du = \frac{dx}{x}$ and substitute $du$:

   $$\int e^{u} \sin{\left(u \right)}\, du$$

   1. Use integration by parts, noting that the integrand eventually repeats itself.

      1. For the integrand $e^{u} \sin{\left(u \right)}$:

         Let $u{\left(u \right)} = \sin{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

         Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - \int e^{u} \cos{\left(u \right)}\, du$.

      2. For the integrand $e^{u} \cos{\left(u \right)}$:

         Let $u{\left(u \right)} = \cos{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

         Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \sin{\left(u \right)}\right)\, du$.

      3. Notice that the integrand has repeated itself, so move it to one side:

         $$2 \int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)}$$

         Therefore,

         $$\int e^{u} \sin{\left(u \right)}\, du = \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}$$

   Now substitute $u$ back in:

   $$\frac{x \sin{\left(\log{\left(x \right)} \right)}}{2} - \frac{x \cos{\left(\log{\left(x \right)} \right)}}{2}$$

2. Now simplify:

   $$- \frac{\sqrt{2} x \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2}$$

3. Add the constant of integration:

   $$- \frac{\sqrt{2} x \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$- \frac{\sqrt{2} x \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2}+ \mathrm{constant}$$