$$\sum_{n=0}^{\infty}\ \int_{\pi/4}^{\pi/3}\sin^{n}x (1-\sin x)^2 dx$$
Let $g_n = \sin^{n}x (1-\sin x)^2$
$g_n$ is a sequence of measurable functions and $g_n \ge 0$ so applying the Beppo Levi Theorem we get -
$$= \int_{\pi/4}^{\pi/3}\sum_{n=0}^{\infty}\sin^{n}x (1-\sin x)^2 dx$$
$$= \int_{\pi/4}^{\pi/3}(1-\sin x)^2 \sum_{n=0}^{\infty}\sin^{n}x dx$$
Now $\sin^{n}x < 1$ for $x \in (\pi/4, \pi/3)$ so we have a geometric series and hence
$$\sum_{n=0}^{\infty}\sin^{n}x = \frac{1}{1 - \sin x}$$
This gives us -
$$= \int_{\pi/4}^{\pi/3}(1-\sin x)^2 \frac{1}{1 - \sin x} dx$$
$$= \int_{\pi/4}^{\pi/3}(1-\sin x)dx$$
$$= x + \cos x \mid_{\pi/4}^{\pi/3}$$
$$\frac{\pi}{3} + \cos \frac{\pi}{3} - \frac{\pi}{4} - \cos \frac{\pi}{4}$$
$$\frac{\pi}{12} + \frac{1}{2} - \frac{1}{\sqrt{2}}$$
Is this correct? In particular, have I got the correct conditions to apply the Beppo Levi Theorem?
The use of Beppo-Levi's theorem and the computations are correct.