Cho hàm số f(x) > 0 có đạo hàm liên tục trên đoạn (0;pi/3)

Ta có \(f''\left( x \right) \cdot f\left( x \right) + {\left[ {\frac{{f\left( x \right)}}{{\cos x}}} \right]^2} = {\left[ {f'\left( x \right)} \right]^2} \Leftrightarrow \frac{{f''\left( x \right) \cdot f\left( x \right) - {{\left[ {f'\left( x \right)} \right]}^2}}}{{{f^2}\left( x \right)}} =  - \frac{1}{{{{\cos }^2}x}}\)

\( \Leftrightarrow \left[ {\frac{{f'\left( x \right)}}{{f\left( x \right)}}} \right] =  - \frac{1}{{{{\cos }^2}x}} \Rightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} =  - \tan x + C.\)

Vì \(\left\{ {\begin{array}{*{20}{l}}{f'\left( 0 \right) = 0}\\{f\left( 0 \right) = 1}\end{array} \Rightarrow C = 0.} \right.\)

Dó đó \(\frac{{f'\left( x \right)}}{{f\left( x \right)}} =  - \tan x \Rightarrow \int\limits_0^{\frac{\pi }{3}} {\frac{{f'\left( x \right)}}{{f\left( x \right)}}}  = \int\limits_0^{\frac{\pi }{3}} {\left( { - \tan x} \right)\;{\rm{d}}x}  \Leftrightarrow \int\limits_0^{\frac{\pi }{3}} {\frac{{d\left( {f\left( x \right)} \right)}}{{f\left( x \right)}}}  = \int\limits_0^{\frac{\pi }{3}} {\frac{{d\left( {\cos x} \right)}}{{\cos x}}} \)

\[\left. { \Leftrightarrow \ln f\left( x \right)} \right|_0^{\frac{\pi }{3}} = \left. {\ln \cos x} \right|_0^{\frac{\pi }{3}} \Leftrightarrow \ln f\left( {\frac{\pi }{3}} \right) - \ln f\left( 0 \right) = \ln \frac{1}{2} - \ln 1 \Leftrightarrow f\left( {\frac{\pi }{3}} \right) = \frac{1}{2}.\]

Vậy \(T = \frac{1}{2}.\) Đáp án: \[\frac{1}{2}\].