求 $\displaystyle \int x \sin^{2} x \, dx$ 。 区域 R 由曲线 $y = \sqrt{x \sin x}$、直线 x = 0 和 $x = \pi$ 以及 x 轴围成。求区域 R 绕 x 轴旋转一周所形成的旋转体的体积。 由此计算区域 S 绕 x 轴旋转一周所形成的旋转体的体积，其中 S 是由曲线 $y = \sqrt{x \sin x}$、直线 $x = \pi$ 和 $y = \sqrt{\pi}$ 以及 y 轴围成的区域。

1. Rewrite the integral using the double-angle identity to eliminate the squared trigonometric term.

$\int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \right) \, dx$
$= \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx$

2. Apply integration by parts to evaluate $\int x \cos 2x \, dx$.

Let $u = x \implies \frac{du}{dx} = 1$, and $\frac{dv}{dx} = \cos 2x \implies v = \frac{1}{2} \sin 2x$.
$\int x \cos 2x \, dx = x \left( \frac{1}{2} \sin 2x \right) - \int 1 \cdot \frac{1}{2} \sin 2x \, dx$
$= \frac{x}{2} \sin 2x - \left( -\frac{1}{4} \cos 2x \right)$
$= \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x$

3. Evaluate the remaining polynomial integral and combine terms with the constant of integration.

$\frac{1}{2} \int x \, dx = \frac{1}{2} \left( \frac{x^2}{2} \right) = \frac{x^2}{4}$
Substituting the pieces back into the expanded expression:
$\int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{1}{2} \left( \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right) + C$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x - \frac{1}{8} \cos 2x + C$

1. Set up the integral for the volume of revolution about the $x$-axis.

$V = \pi \int_{a}^{b} y^2 \, dx$
$V = \pi \int_{0}^{\pi} (\sqrt{x \sin x})^2 \, dx = \pi \int_{0}^{\pi} x \sin x \, dx$

2. Evaluate the indefinite integral $\int x \sin x \, dx$ using integration by parts.

Let $u = x \implies \frac{du}{dx} = 1$
Let $\frac{dv}{dx} = \sin x \implies v = -\cos x$

$\int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx$
$= -x \cos x + \sin x$

3. Apply the limits of integration to calculate the exact volume.

$V = \pi \left[ -x \cos x + \sin x \right]_{0}^{\pi}$
$V = \pi \Big( (-\pi \cos \pi + \sin \pi) - (-0 \cos 0 + \sin 0) \Big)$
$V = \pi \Big( (-\pi(-1) + 0) - (0 + 0) \Big)$
$V = \pi^2$

1. Calculate the volume of revolution of region $R$ from part (ii), as it is needed to find the volume of $S$.

Since the target relies on part (ii), we first calculate the volume of $R$ rotated about the $x$-axis.
$V_R = \pi \int_0^\pi (\sqrt{x \sin x})^2 \, dx = \pi \int_0^\pi x \sin x \, dx$

Using integration by parts, let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$:
$\int_0^\pi x \sin x \, dx = \left[ -x \cos x \right]_0^\pi - \int_0^\pi (-\cos x) \, dx$
$= (-\pi \cos \pi - 0) + \left[ \sin x \right]_0^\pi = \pi(1) + 0 = \pi$

Thus, $V_R = \pi(\pi) = \pi^2$.

2. Formulate the integral for the volume of the solid generated by rotating region $S$.

Region $S$ is bounded above by $y = \sqrt{\pi}$, below by the curve $y = \sqrt{x \sin x}$, on the left by $x = 0$ (the $y$-axis), and on the right by $x = \pi$.

When rotated through 4 right angles ($360^\circ$ or $2\pi$ radians) about the $x$-axis, the volume $V_S$ can be found using the washer method (outer volume minus inner volume):
$V_S = \pi \int_0^\pi \left( (\sqrt{\pi})^2 - (\sqrt{x \sin x})^2 \right) \, dx$
$V_S = \pi \int_0^\pi (\pi - x \sin x) \, dx$

3. Evaluate the integral to find $V_S$.

Expanding the integral and substituting $V_R$:
$V_S = \pi \int_0^\pi \pi \, dx - \pi \int_0^\pi x \sin x \, dx$
$V_S = \pi \left[ \pi x \right]_0^\pi - V_R$
$V_S = \pi (\pi^2 - 0) - \pi^2$
$V_S = \pi^3 - \pi^2$