Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $f\left(x \right)+f\left(\dfrac{\pi }{2}-x \right)=\sin x.\cos...

Thay $x=\dfrac{\pi }{2}$ vào đẳng thức $f\left( x \right)+f\left( \dfrac{\pi }{2}-x \right)=\sin x.\cos x\Rightarrow f\left( \dfrac{\pi }{2} \right)+f\left( 0 \right)=0\Leftrightarrow f\left( \dfrac{\pi }{2} \right)=0.$
Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{x.f'\left( x \right)}dx$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\Rightarrow I=x.f\left( x \right)\left| _{0}^{\dfrac{\pi }{2}} \right.-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}\left( 1 \right)$
Lại có: $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\cos xdx}$
$\Leftrightarrow 2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2xdx}\Leftrightarrow 2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\dfrac{1}{2}\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\dfrac{1}{4}.$
Vậy $I=-\dfrac{1}{4}.$