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(D) Given $f(x) + \int_0^x f(t) \sqrt{1 - (\ln f(t))^2} dt = e$. At $x=0$,$f(0) + 0 = e$,so $f(0) = e$. Differentiating both sides with respect to $x$

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$27$

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(D) Given $f(x) + \int_0^x f(t) \sqrt{1 - (\ln f(t))^2} dt = e$. At $x=0$,$f(0) + 0 = e$,so $f(0) = e$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0$. Let $y = f(x)$,then $\frac{dy}{dx} = -y \sqrt{1 - (\ln y)^2}$. Separating variables: $\int \frac{dy}{y \sqrt{1 - (\ln y)^2}} = -\int dx$. Let $\ln y = t$,then $\frac{1}{y} dy = dt$. $\int \frac{dt}{\sqrt{1 - t^2}} = -x + C$. $\sin^{-1}(t) = -x + C \Rightarrow \sin^{-1}(\ln f(x)) = -x + C$. Since $f(0) = e$,$\sin^{-1}(\ln e) = -0 + C \Rightarrow \sin^{-1}(1) = C \Rightarrow C = \frac{\pi}{2}$. Thus,$\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x$. For $x = \frac{\pi}{6}$,$\sin^{-1}(\ln f(\frac{\pi}{6})) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. Therefore,$\ln f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$. Finally,$(6 \ln f(\frac{\pi}{6}))^2 = (6 	imes \frac{\sqrt{3}}{2})^2 = (3\sqrt{3})^2 = 27$.

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