(Chuyên Lam Sơn 2022) Cho hàm số \(y = f(x)\) có đạo hàm liên tục trên \((0;\pi )\) thỏa mãn \(f\prime (x) = f(x)\). \(\cot x + 2x\). \(\sin x\). Biết \(f\left( {\frac{\pi }{2}} \right) = \frac{{{\pi ^2}}}{4}\). Tính \(f\left( {\frac{\pi }{6}} \right)\).

A. \(\frac{{{\pi ^2}}}{{36}}\).

B. \(\frac{{{\pi ^2}}}{{72}}\).

C. \(\frac{{{\pi ^2}}}{{54}}\).

D. \(\frac{{{\pi ^2}}}{{80}}\).

Lời giải:

\(\begin{array}{l}f\prime (x) = f(x) \cdot \cot x + 2x \cdot \sin x \Leftrightarrow \sin x \cdot f\prime (x) – f(x) \cdot \cos x + 2x \cdot {\sin ^2}x\\ \Leftrightarrow \sin x \cdot f\prime (x) – f(x) \cdot \cos x = 2x \cdot {\sin ^2}x\\ \Leftrightarrow \frac{{\sin x \cdot f\prime (x) – f(x) \cdot \cos x}}{{{{\sin }^2}x}} = 2x\\ \Rightarrow \int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\sin x \cdot f\prime (x) – f(x) \cdot \cos x}}{{{{\sin }^2}x}}} dx = \int_{\frac{\pi }{6}}^{\frac{\pi }{2}} 2 x \cdot dx\\ \Leftrightarrow \int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\left( {\frac{{f(x)}}{{\sin x}}} \right)} dx = \left. {{x^2}} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}}\left. { \Leftrightarrow \frac{{f(x)}}{{\sin x}}} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}} = \frac{{{\pi ^2}}}{4} – \frac{{{\pi ^2}}}{{36}}\\ \Leftrightarrow \frac{{f\left( {\frac{\pi }{2}} \right)}}{1} – \frac{{f\left( {\frac{\pi }{6}} \right)}}{{\frac{1}{2}}} = \frac{{{\pi ^2}}}{4} – \frac{{{\pi ^2}}}{{36}} \Leftrightarrow f\left( {\frac{\pi }{6}} \right) = \frac{{{\pi ^2}}}{{72}}\end{array}\)