2022-M2-10

$\begin{array}{cl}& {\displaystyle {\int }_0^{\pi }g(x)dx}\\ =& {\left[{\displaystyle \frac{\sin 2x{\cos }^{2}x}{2}}\right]}_0^{\pi }+{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi }{\sin }^{2}2xdx}\\ =& {\displaystyle \frac{\sin 2\pi {\cos }^2\pi }{2}}-{\displaystyle \frac{\sin 2(0){\cos }^{2}0}{2}}+{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi }{\displaystyle \frac{1-\cos 4x}{2}}dx}\\ =& {\displaystyle \frac{1}{4}}{[x-{\displaystyle \frac{\sin 4x}{4}}]}_0^{\pi }\\ =& {\displaystyle \frac{1}{4}}(\pi -{\displaystyle \frac{\sin 4\pi }{4}})-{\displaystyle \frac{1}{4}}(0-{\displaystyle \frac{\sin 4(0)}{4}})\\ =& {\displaystyle \frac{\pi }{4}}\end{array}$

當 $x=0$ 時，$u=\pi$。

當 $x=\pi$ 時，$u=0$。

$\begin{array}{rcl}{\displaystyle {\int }_0^{\pi }xg(x)dx}& =& {\displaystyle {\int }_{\pi }^0(\pi -u){\cos }^2(\pi -u)\cos 2(\pi -u)(-du)}\\ {\displaystyle {\int }_0^{\pi }xg(x)dx}& =& {\displaystyle {\int }_0^{\pi }(\pi -u){\cos }^{2}u\cos 2udu}\\ {\displaystyle {\int }_0^{\pi }xg(x)dx}& =& \pi {\displaystyle {\int }_0^{\pi }g(u)du-{\displaystyle {\int }_0^{\pi }ug(u)du}}\\ {\displaystyle {\int }_0^{\pi }xg(x)dx}& =& \pi {\displaystyle {\int }_0^{\pi }g(x)dx-{\displaystyle {\int }_0^{\pi }xg(x)dx\text{ , 由於 }u\text{ 為一啞變數。}}}\\ 2{\displaystyle {\int }_0^{\pi }xg(x)dx}& =& \pi {\displaystyle {\int }_0^{\pi }g(x)dx}\\ {\displaystyle {\int }_0^{\pi }xg(x)dx}& =& {\displaystyle \frac{1}{2}}\pi \left({\displaystyle \frac{\pi }{4}}\right)\text{ , 利用 (b) 的結果。}\\ {\displaystyle {\int }_0^{\pi }xg(x)dx}& =& {\displaystyle \frac{{\pi }^2}{8}}\end{array}$

考慮右方的第一項，

$\begin{array}{rcl}(-x)g(-x)& =& (-x){\cos }^2(-x)\cos 2(-x)\\ (-x)g(-x)& =& -x{\cos }^{2}x\cos 2x\\ (-x)g(-x)& =& -xg(x)\end{array}$

$\therefore xg(x)$ 為一奇函數。

所以，${\displaystyle {\int }_{-\pi }^{\pi }xg(x)dx=0}$。

考慮右方的第二項，設 $u=x-\pi$。則 $du=dx$。

當 $x=\pi$ 時，$u=0$。

當 $x=2\pi$ 時，$u=\pi$。

$\begin{array}{cl}& {\displaystyle {\int }_{\pi }^{2\pi }xg(x)dx}\\ =& {\displaystyle {\int }_0^{\pi }(\pi +u){\cos }^2(\pi +u)\cos 2(\pi +u)du}\\ =& {\displaystyle {\int }_0^{\pi }(\pi +u){\cos }^{2}u\cos 2udu}\\ =& \pi {\displaystyle {\int }_0^{\pi }{\cos }^{2}u\cos 2udu+{\displaystyle {\int }_0^{\pi }u{\cos }^{2}u\cos 2udu}}\\ =& \pi {\displaystyle {\int }_0^{\pi }{\cos }^{2}x\cos 2xdx+{\displaystyle {\int }_0^{\pi }x{\cos }^{2}x\cos 2xdx\text{ , 由於 }u\text{ 為一啞變數。}}}\\ =& \pi {\displaystyle {\int }_0^{\pi }g(x)dx+{\displaystyle {\int }_0^{\pi }xg(x)dx}}\\ =& \pi \times {\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{{\pi }^2}{8}}\text{ , 利用 (b) 及 (c) 的結果。}\\ =& {\displaystyle \frac{3{\pi }^2}{8}}\end{array}$

由此，可得

$\begin{array}{rcl}{\displaystyle {\int }_{-\pi }^{2\pi }xg(x)dx}& =& {\displaystyle {\int }_{-\pi }^{\pi }xg(x)dx+{\displaystyle {\int }_{\pi }^{2\pi }xg(x)dx}}\\ {\displaystyle {\int }_{-\pi }^{2\pi }xg(x)dx}& =& 0+{\displaystyle \frac{3{\pi }^2}{8}}\\ {\displaystyle {\int }_{-\pi }^{2\pi }xg(x)dx}& =& {\displaystyle \frac{3{\pi }^2}{8}}\end{array}$