Lots of Sines

By way of the Monotone Convergence Theorem , we can write the sum of integrals as

$I = \displaystyle\sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \sin^{n}(x) dx = \int_{0}^{\frac{\pi}{4}} \sum_{n=1}^{\infty} \sin^{n}(x) dx.$

Now for any $x$ in $[0, \frac{\pi}{4}]$ we have that $|\sin(x)| \lt 1$ , and thus

$\displaystyle\sum_{n=1}^{\infty} \sin^{n}(x) = \dfrac{\sin(x)}{1 - \sin(x)} = \dfrac{\sin(x)}{1 - \sin(x)} * \dfrac{1 + \sin(x)}{1 + \sin(x)} =$

$\dfrac{\sin(x) + \sin^{2}(x)}{\cos^{2}(x)} = \dfrac{\sin(x)}{\cos^{2}(x)} + \dfrac{1 - \cos^{2}(x)}{\cos^{2}(x)} = \dfrac{\sin(x)}{\cos^{2}(x)} + \sec^{2}(x) - 1.$

The indefinite integral of this expression (without the constant) is

$\sec(x) + \tan(x) - x,$

where the substitution $u = \cos(x)$ was used to evaluate the integral of the first term. Evaluating this last expression from $x = 0$ to $x = \frac{\pi}{4}$ gives us the solution

$I = (\sqrt{2} + 1 - \dfrac{\pi}{4}) - (1 + 0 - 0) = \sqrt{2} - \dfrac{\pi}{4} = 0.6288$

to 4 decimal places, and thus $\lfloor 1000I \rfloor = \boxed{628}.$