$\begin{array}{l}
\int\limits_0^{\dfrac{\pi }{2}} {\sin 2xf'\left( {\sin x} \right)dx} \\
 = \int\limits_0^{\dfrac{\pi }{2}} {2\sin x\cos xf'\left( {\sin x} \right)dx} \\
 = 2\int\limits_0^{\dfrac{\pi }{2}} {\sin xf'\left( {\sin x} \right)d\left( {\sin x} \right)}  = 2{I_1}\\
\left\{ \begin{array}{l}
\sin x = u\\
f'\left( {\sin x} \right)d\left( {\sin x} \right) = dv
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\cos xdx = du\\
f\left( {\sin x} \right) = v
\end{array} \right.\\
 \Rightarrow {I_1} = \left. {\sin xf\left( {\sin x} \right)} \right|_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos xdx} \\
 = \sin \dfrac{\pi }{2}f\left( {\sin \dfrac{\pi }{2}} \right) - \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos xdx} \\
 = f\left( 1 \right) - \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos xdx} \\
 = 1 - \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos xdx} \\
\sin x = t,x = 0 \to t = 0,x = \dfrac{\pi }{2} \to t = 1\\
 \Rightarrow \cos xdx = dt\\
 \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos xdx}  = \int\limits_0^1 {f\left( t \right)dt}  = \dfrac{1}{3}\\
 \Rightarrow {I_1} = 1 - \dfrac{1}{3} = \dfrac{2}{3}
\end{array}$