Let $f$ be $a$ differentiable function defined on $\left[ {0,{\pi \over 2}} \right]$ such that $f(x) > 0$ and $$f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}$$. Then $\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2$ is equal to __________.

$f(x)+\int_{0}^{x} f(t) \sqrt{1-\left(\log _{e} f(t)\right)^{2}} d t=e\quad...(1)$ <br/><br/> So, $f(0)=e$ <br/><br/> Now differentiate w.r. to $x$ <br/><br/> $$ \begin{gathered} f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\ \frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\right)^{2}}}=-1 \end{gathered} $$ <br/><br/> Let $\log _{e} f(x)=t$ <br/><br/> $\therefore \int \frac{d t}{\sqrt{1-t^{2}}}=-x+c$ <br/><br/> $\Rightarrow \sin ^{-1} t=-x+c$ <br/><br/> Now $f(0)=e \Rightarrow t=1$ So, $c=\frac{\pi}{2}$ <br/><br/> $\therefore t=\sin \left(\frac{\pi}{2}-x\right)=\cos x \quad\left(\because x \in\left[0, \frac{\pi}{2}\right]\right)$ <br/><br/> $\therefore\left(6 \log _{e} f\left(\frac{\pi}{6}\right)\right)^{2}=27$