2022-M2-10

$\begin{array}{cl}
& \dint_0^\pi g(x) dx \\
= & \left[\dfrac{\sin 2x \cos^2 x}{2}\right]_0^\pi +\dfrac{1}{2}\dint_0^\pi \sin^2 2xdx \\
= & \dfrac{\sin2\pi\cos^2 \pi}{2}-\dfrac{\sin2(0)\cos^2 0}{2} +\dfrac{1}{2} \dint_0^\pi \dfrac{1-\cos 4x}{2} dx \\
= & \dfrac{1}{4}\left[x-\dfrac{\sin 4x}{4}\right]_0^\pi \\
= & \dfrac{1}{4}\left(\pi-\dfrac{\sin4\pi}{4}\right)-\dfrac{1}{4}\left(0-\dfrac{\sin4(0)}{4}\right) \\
= & \dfrac{\pi}{4}
\end{array}$

When $x=0$, $u=\pi$.

When $x=\pi$, $u=0$.

$\begin{array}{rcl}
\dint_0^\pi xg(x) dx & = & \dint_\pi^0 (\pi-u)\cos^2(\pi-u)\cos2(\pi-u) (-du) \\
\dint_0^\pi xg(x) dx & = & \dint_0^\pi (\pi-u)\cos^2 u\cos 2u du \\
\dint_0^\pi xg(x) dx & = & \pi\dint_0^\pi g(u)du-\dint_0^\pi ug(u)du \\
\dint_0^\pi xg(x) dx & = & \pi\dint_0^\pi g(x)dx-\dint_0^\pi xg(x)dx \text{ , since $u$ is a dummy variable.} \\
2\dint_0^\pi xg(x) dx & = & \pi\dint_0^\pi g(x) dx \\
\dint_0^\pi xg(x) dx & = & \dfrac{1}{2}\pi \left(\dfrac{\pi}{4}\right) \text{ , by the result of (b).} \\
\dint_0^\pi xg(x) dx & = & \dfrac{\pi^2}{8}
\end{array}$

For the first term of the right side,

$\begin{array}{rcl}
(-x)g(-x) & = & (-x)\cos^2 (-x)\cos2(-x) \\
(-x)g(-x) & = & -x\cos^2 x \cos2x \\
(-x)g(-x) & = & -xg(x)
\end{array}$

$\therefore xg(x)$ is an odd function.

Hence, we have $\dint_{-\pi}^\pi xg(x)dx=0$.

For the second term of the right side, let $u=x-\pi$. Then $du=dx$.

When $x=\pi$, $u=0$.

When $x=2\pi$, $u=\pi$.

$\begin{array}{cl}
& \dint_\pi^{2\pi} xg(x) dx \\ = & \dint_0^\pi(\pi+u)\cos^2(\pi+u)\cos2(\pi+u)du \\
= & \dint_0^\pi (\pi+u) \cos^2 u\cos 2u du \\
= & \pi\dint_0^\pi\cos^2u\cos 2udu +\dint_0^\pi u\cos^2 u\cos2u du \\
= & \pi \dint_0^\pi \cos^2 x \cos 2x dx + \dint_0^\pi x\cos^2 x\cos 2x dx \text{ , since $u$ is a dummy variable.} \\
= & \pi\dint_0^\pi g(x)dx +\dint_0^\pi xg(x)dx \\
= & \pi\times \dfrac{\pi}{4}+\dfrac{\pi^2}{8} \text{ , by the result of (b) and (c).} \\
= & \dfrac{3\pi^2}{8}
\end{array}$

Hence, we have

$\begin{array}{rcl}
\dint_{-\pi}^{2\pi} xg(x) dx & = & \dint_{-\pi}^\pi xg(x) dx +\dint_\pi^{2\pi}xg(x) dx \\
\dint_{-\pi}^{2\pi} xg(x) dx & = & 0 + \dfrac{3\pi^2}{8} \\
\dint_{-\pi}^{2\pi} xg(x) dx & = & \dfrac{3\pi^2}{8}
\end{array}$