Evaluate $I = \displaystyle\int_{e^{\pi/6}}^{e^{\pi/2}} \frac{\sin(\ln(\sin(\ln x)))\cdot\cos(\ln x)}{x\sin(\ln x)}\,dx$. Find $\cos^{-1}(I+1)$.
Step-by-Step Solution
Key Concept: Substitute $t = \ln x$, then $u = \sin t$; integral becomes $\int\frac{\sin(\ln u)}{u}$ which further simplifies.
Let $t=\ln x$: $I=\int_{\pi/6}^{\pi/2}\frac{\sin(\ln\sin t)\cos t}{\sin t}dt$. Let $u=\sin t$: $I=\int_{1/2}^{1}\frac{\sin(\ln u)}{u}du$. Let $v=\ln u$: $I=\int_{-\ln 2}^{0}\sin v\,dv = [-\cos v]_{-\ln 2}^{0}=-(1-\cos(\ln 2))=\cos(\ln 2)-1$. So $\cos^{-1}(I+1)=\cos^{-1}(\cos(\ln 2))=\ln 2$.
Correct Answer: C