\(=\int\limits^{\dfrac{\pi}{2}}_0x\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)dx=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{x}{2}dx-\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0x.cos2xdx=\dfrac{\pi^2}{16}-\dfrac{1}{2}I_1\)
Với \(I_1=\int\limits^{\dfrac{\pi}{2}}_0x.cos2xdx\)
Đặt \(\left\{{}\begin{matrix}u=x\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)
\(\Rightarrow I_1=\dfrac{1}{2}x.sin2x|^{\dfrac{\pi}{2}}_0-\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0sin2xdx=0-\dfrac{1}{2}=-\dfrac{1}{2}\)
\(\Rightarrow I=\dfrac{\pi^2}{16}-\dfrac{1}{2}\left(-\dfrac{1}{2}\right)=\dfrac{\pi^2}{16}+\dfrac{1}{4}\)
