Evaluation of

$$\int_{\theta = 0}^\pi \sin^3 \theta d \theta~:$$

$$\int_0^\pi \sin^3\theta d\theta = \int_0^\pi \sin\theta (1 - \cos^2\theta) d \theta$$
$$\int_0^\pi \sin^3\theta d\theta = \int_0^\pi \sin\theta d \theta + \int_0^\pi \cos^2\theta (-\sin\theta d\theta)$$ Let $x \equiv \cos\theta$ so $dx = -\sin\theta d\theta$.
$$\int_0^\pi \sin^3\theta d\theta = \cos\theta \vert_\pi^0 + \int_{1}^{-1} x^2 dx$$
$$ \int_0^\pi \sin^3\theta d\theta = 2 + {x^3 \over 3}\bigg|_1^{-1} = 2 - 1/3 - 1/3 = 4/3$$