Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và thỏa mãn $f\left( -x \right)+2021f\left( x \right)=x\sin ,\forall x\in \mathbb{R}$...

Từ giả thuyết: $f\left( -x \right)+2021f\left( x \right)+x\sin x,\forall x\in \mathbb{R}$
$\Leftrightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( -x \right)dx}+2021\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{x\sin xdx\left( * \right)}$
Tính: $\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( -x \right)dx}\overset{t=-x}{\mathop{=}} -\int\limits_{\dfrac{\pi }{2}}^{-\dfrac{\pi }{2}}{f\left( t \right)dt}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( t \right)dt}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=I.$
Tính: $\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{x\sin xdx}$. Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=\sin xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=-\cos x \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{x\sin xdx}=-x\cos x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& -\dfrac{\pi }{2} \\
\end{aligned} \right.+\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\cos xdx}=\sin x\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& -\dfrac{\pi }{2} \\
\end{aligned} \right.=2$
$\left( * \right)\Leftrightarrow I+2021.I=2\Leftrightarrow I=\dfrac{1}{1011}$.