Cho hàm số f(x) liên tục trên đoạn \[\left[ {\frac{\pi }{4};\frac{\pi }{3}} \right]\] thỏa mãn \[f'\left( x \right).\sin 2x = 1 + 2.f\left( x \right)\] với \[\forall x \in \left[ {\frac{\pi }{4};\frac{\pi }{3}} \right]\] và \[f\left( {\frac{\pi }{4}} \right) = 1.\] Tích phân \[I = \int\limits_0^{\frac{\pi }{3}} {f\left( x \right)dx} \] bằng

Đáp án A

Ta có

\({\left[ {\frac{{f\left( x \right)}}{{\tan x}}} \right]^\prime } = \frac{{f'\left( x \right).\tan x - f\left( x \right).\frac{1}{{{{\cos }^2}x}}}}{{{{\tan }^2}x}} = \frac{{f'\left( x \right).\sin x\cos x - f\left( x \right)}}{{{{\sin }^2}x}}\)

\( = \frac{{f'\left( x \right).\sin 2x - 2.f\left( x \right)}}{{2{{\sin }^2}x}} = \frac{1}{{2{{\sin }^2}x}} \Rightarrow \frac{{f\left( x \right)}}{{\tan x}} = \mathop \smallint \nolimits^ \frac{1}{{2{{\sin }^2}x}}dx = - \frac{1}{2}\cot x + C.\)

Bài ra

\(f\left( {\frac{\pi }{4}} \right) = 1 \Rightarrow C = \frac{3}{2} \Rightarrow f\left( x \right) = \tan x\left( { - \frac{1}{2}\cot x + \frac{3}{2}} \right) = - \frac{1}{2} + \frac{3}{2}.\frac{{\sin x}}{{\cos x}}\)

\( \Rightarrow \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {f\left( x \right)dx} = \left( { - \frac{1}{2}x - \frac{3}{2}\ln \left| {\cos x} \right|} \right)\left| {_{\scriptstyle\atop\scriptstyle\frac{\pi }{4}}^{\scriptstyle\frac{\pi }{3}\atop\scriptstyle}} \right. = - \frac{\pi }{6} - \frac{3}{2}\ln \frac{1}{2} + \frac{\pi }{8} + \frac{3}{2}\ln \frac{1}{{\sqrt 2 }} = - \frac{\pi } + \frac{3}{4}\ln 2.\)