Đáp án đúng: D
Phương pháp giải:
Lấy tích phân từ \(0\) đến \(\dfrac{\pi }{2}\) hai vế phương trình \(\sin x\,f\left( {\cos x} \right) + \cos x\,f\left( {\sin x} \right) = \sin 2x - \frac{1}{2}{\sin ^3}2x\).
Giải chi tiết:Lấy tích phân từ \(0\) đến \(\dfrac{\pi }{2}\) hai vế phương trình \(\sin x\,f\left( {\cos x} \right) + \cos x\,f\left( {\sin x} \right) = \sin 2x - \dfrac{1}{2}{\sin ^3}2x\) ta được:
\(\begin{array}{l}\,\,\,\,\,\,\int\limits_0^{\frac{\pi }{2}} {\left[ {\sin x\,f\left( {\cos x} \right) + \cos x\,f\left( {\sin x} \right)} \right]} \,dx = \int\limits_0^{\frac{\pi }{2}} {\left[ {\sin 2x - \frac{1}{2}{{\sin }^3}2x} \right]} \,dx\\ \Leftrightarrow \int\limits_0^{\frac{\pi }{2}} {\sin x\,f\left( {\cos x} \right)} \,dx + \int\limits_0^{\frac{\pi }{2}} {\cos x\,f\left( {\sin x} \right)} \,dx = \int\limits_0^{\frac{\pi }{2}} {\left[ {\sin 2x - \frac{1}{2}.\frac{{3\sin 2x - \sin 6x}}{4}} \right]} \,dx\\ \Leftrightarrow - \int\limits_0^{\frac{\pi }{2}} {\,f\left( {\cos x} \right)} \,d\left( {\cos x} \right) + \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)} \,d\left( {\sin x} \right) = \left. {\left( { - \frac{1}{2}\cos 2x + \frac{3}{{16}}\cos 2x - \frac{1}{{48}}\cos 6x} \right)} \right|_0^{\frac{\pi }{2}}\\ \Rightarrow - \int\limits_1^0 {\,f\left( t \right)} \,dt + \int\limits_0^1 {f\left( u \right)} \,du = \left( {\frac{1}{2} - \frac{3}{{16}} + \frac{1}{{48}}} \right) - \left( { - \frac{1}{2} + \frac{3}{{16}} - \frac{1}{{48}}} \right)\\ \Leftrightarrow \int\limits_0^1 {\,f\left( x \right)} \,dx + \int\limits_0^1 {f\left( x \right)} \,dx = \frac{2}{3} \Leftrightarrow \int\limits_0^1 {f\left( x \right)} \,dx = \frac{1}{3}\end{array}\)
Chọn D.
