evaluate using integration by parts, substitution, or both if necessary.\n(use symbolic notation and…

Explanation:

Step1: Substitution

Let (u = \ln(2x)), then (x=e^{u}) and (dx = e^{u}du). The integral becomes (\int\sin(u)e^{u}du).

Step2: Integration by Parts

Let (v=\sin(u)), (dw = e^{u}du). Then (dv=\cos(u)du), (w = e^{u}). By integration - by - parts formula (\int vdw=vw-\int wdv), we have (\int\sin(u)e^{u}du=e^{u}\sin(u)-\int e^{u}\cos(u)du).

Step3: Integration by Parts again

For (\int e^{u}\cos(u)du), let (v = \cos(u)), (dw=e^{u}du). Then (dv=-\sin(u)du), (w = e^{u}). (\int e^{u}\cos(u)du=e^{u}\cos(u)+\int e^{u}\sin(u)du).

Step4: Solve for (\int\sin(u)e^{u}du)

Substitute the result of (\int e^{u}\cos(u)du) into the equation from Step2: (\int\sin(u)e^{u}du=e^{u}\sin(u)-\left(e^{u}\cos(u)+\int e^{u}\sin(u)du\right)) (2\int\sin(u)e^{u}du=e^{u}(\sin(u)-\cos(u))) (\int\sin(u)e^{u}du=\frac{1}{2}e^{u}(\sin(u)-\cos(u))+C)

Step5: Back - substitution

Since (u = \ln(2x)), (e^{u}=2x). (\int\sin(\ln(2x))dx=\frac{1}{2}x(\sin(\ln(2x))-\cos(\ln(2x)))+C)

Answer:

(\frac{1}{2}x(\sin(\ln(2x))-\cos(\ln(2x)))+C)