Question:

Step 1: Apply the substitution $t = \ln x$ to simplify the integral. Let $t = \ln x$, which gives us $dt = \frac{dx}{x}$. When $x = e^{\pi/6}$, we have $t = \pi/6$ When $x = e^{\pi/2}$, we have $t = \pi/2$ The integral becomes: $$I = \int_{\pi/6}^{\pi/2} \frac{\sin(\ln(\sin t))\cos t}{\sin t}\, dt$$ Step 2: Apply a second substitution to further simplify. Let $u = \ln(\sin t)$, so: $$du = \frac{\cos t}{\sin t}\, dt = \cot t\, dt$$ We need to find the new limits of integration: When $t = \pi/6$: $u = \ln\left(\sin\frac{\pi}{6}\right) = \ln\left(\frac{1}{2}\right) = -\ln 2$ When $t = \pi/2$: $u = \ln(\sin(\pi/2)) = \ln(1) = 0$ Step 3: Evaluate the transformed integral. The integral now becomes: $$I = \int_{-\ln 2}^{0} \sin(u)\, du$$ Evaluating this standard integral: $$I = \left[-\cos u\right]_{-\ln 2}^{0} = -\cos(0) - \left(-\cos(-\ln 2)\right)$$ Since $\cos(-\ln 2) = \cos(\ln 2)$ (cosine is an even function): $$I = -1 + \cos(\ln 2)$$ Step 4: Find $I + 1$ and apply the inverse cosine. $$I + 1 = -1 + \cos(\ln 2) + 1 = \cos(\ln 2)$$ Therefore: $$\cos^{-1}(I+1) = \cos^{-1}(\cos(\ln 2)) = \ln 2$$ Since $\ln 2 \approx 0.693$ lies in the range $[0, \pi]$ of the inverse cosine function, we have: $$\boxed{\cos^{-1}(I+1) = \ln 2}$$ The answer is **Option 3: $\ln 2$**