First a correction:
$$\begin{align}
\frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) & =-\alpha \sin \theta \, e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\
\end{align}$$
Now
\begin{align}
\frac{dI}{d\alpha}&=\frac{d}{d\alpha}\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta) d\theta \\
&=\int_{0}^{2\pi}\frac{d}{d\alpha}(e^{\alpha \cos \theta}\cos(\alpha \sin \theta)) d\theta \\
&=\int_{0}^{2\pi}\cos \theta \, e^{\alpha \cos \theta}\cos(\alpha \sin \theta)- e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\sin \theta \, d\theta \\
&=\int_{0}^{2\pi}\frac{1}{\alpha} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) d\theta \\
&=\frac{1}{\alpha} \Big[e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\Big]_0^{2\pi} \\
&=0
\end{align}
So $I(\alpha)$ is actually constant.
So $I(1)=I(0)=2\pi$
So the answer is $2\pi$
It is actually very easy to express the integral as an infinite series. Note that
$$\sqrt{1 + x} = 1 + \sum_{k = 1}^\infty c_k x^k$$
$$c_k = (-1)^{k+1}\frac{(2k - 3)!!}{k! \cdot 2^k}$$
and
$$\sin x \sin 2x = 2\sin^2 x \cos x$$
Then
$$
\begin{aligned}I &= \int_{\pi/2}^\pi\sqrt{1 + \sin x \sin 2x - \frac{1}{2}\cos^2 x} \;\mathrm{d}x\\
&= \int_{\pi/2}^\pi 1 + \sum_{k = 1}^\infty c_k \left(2\sin^2 x \cos x - \frac{1}{2} \cos^2 x\right)^k \;\mathrm{d}x\\
&= \frac{\pi}{2} + \sum_{k = 1}^\infty c_k \int_0^{\pi/2} \left(-2\cos^2 x \sin x - \frac{1}{2} \sin^2 x\right)^k \;\mathrm{d}x\\
&= \frac{\pi}{2} + \sum_{k = 1}^\infty (-1)^k c_k \int_0^{\pi/2} \sum_{m = 0}^k \binom{k}{m}\left(2\cos^2 x\sin x\right)^m\left(\frac{1}{2}\sin^2 x\right)^{k - m} \;\mathrm{d}x
\end{aligned}$$
Now we have
$$\left(2\cos^2 x\sin x\right)^m\left(\frac{1}{2}\sin^2 x\right)^{k - m} = 2^{2m - k} \sin^{2k - m} x \cos^{2m} x$$
Table of Integrals, Series, and Products (8th edition, p. 397, formula 3.621-5) tells me that
$$\int_0^{\pi/2} \sin^m x \cos^n x \;\mathrm{d}x = \frac{1}{2}\mathrm{B}\left(\frac{m + 1}{2},\frac{n + 1}{2}\right)$$
where $\mathrm{B}(a,b)$ is the complete Beta function. Therefore
$$I = \frac{\pi}{2} + \sum_{k = 1}^\infty (-1)^k c_k \sum_{m = 0}^k \binom{k}{m} 2^{2m - k - 1} \mathrm{B}\left(\frac{2k-m+1}{2},\frac{2m+1}{2}\right)$$
The numerical results seem to agree, but the series converges very slowly. The first 500 terms sum up to 0.82776000293935595965, which still has an error of $10^{-12}$.
Because this is a double sum, we need $O(n^2)$ time to calculate the first $n$ terms. I have not figured out a faster way to calculate.
Update: This problem oddly reminds me of a binomial sum problem on ProjectEuler (Problem 567). Note that the Beta function satisfies
$$\mathrm{B}(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)}$$
For multiples of $1/2$, we can compute the Gamma function explicitly
$$\Gamma\left(\frac{n}{2}\right) = \left\{\begin{aligned}
\left(\frac{n}{2} - 1\right)! & \quad n \text{ even}\\
\sqrt{\pi}\frac{(n - 2)!!}{2^\frac{n - 1}{2}} & \quad n \text{ odd}\end{aligned}\right.$$
Let us rewrite the integral as
$$I = \frac{\pi}{2} - \sum_{k = 1}^\infty \sum_{m = 0}^k d_{k,m}$$
$$d_{k,m} = \frac{(2k - 3)!!}{k! \cdot 2^k} \binom{k}{m} 2^{2m - k - 1} \mathrm{B}\left(\frac{2k-m+1}{2},\frac{2m+1}{2}\right)$$
The calculation of the Beta function term will depend on the parity of $m$. It is easy to verify that
$$\frac{d_{k,2t + 2}}{d_{k,2t}} = \frac{4(k - 2t - 1)(k - 2t)(4t + 1)(4t + 3)}{(2k - 2t - 1)(k + t + 1)(t + 1)(2t + 1)}$$
$$\frac{d_{k,2t + 3}}{d_{k,2t + 1}} = \frac{4(k - 2t - 2)(k - 2t - 1)(4t + 3)(4t + 5)}{(k - t - 1)(2k + 2t + 3)(t + 1)(2t + 3)}$$
$$\frac{d_{k + 1,2t}}{d_{k,2t}} = \frac{(2k - 2t + 1)(2k - 1)}{8(k - 2t + 1)(k + t + 1)}$$
$$\frac{d_{k + 1,2t + 1}}{d_{k,2t + 1}} = \frac{(2k - 1)(k - t)}{2(k - 2t)(2k + 2t + 3)}$$
From here I think it is possible to transform the double sum into a recurrence relation, using hypergeometric summation techniques. I will post results if I have time to do it.
Update: I tried Zeilberger's algorithm. It works but the results look very messy, and is probably not a good idea to get the numerical value of this series. Since Zeilberger's algorithm is a late 20th century invention, I do not think the original problem poser has a better way to calculate the integral, other than adding the series by hand. See Zeilberger's book (https://www.math.upenn.edu/~wilf/AeqB.html) for a description of the algorithm if you want to repeat the attempt.
Update again: I am very impressed by the numerical capabilities of Mathematica. It turns out that we can simply compute
$$I = E + O$$
$$E = \sum_{t = 0}^\infty d_{2t,2t} \cdot {}_2F_1\left(t + \frac{1}{2},2t - \frac{1}{2};3t + 1;\frac{1}{2}\right)$$
$$O = \sum_{t = 0}^\infty d_{2t + 1,2t + 1} \cdot {}_2F_1\left(t + 1,2t + \frac{1}{2};3t + \frac{5}{2};\frac{1}{2}\right)$$
Using Mathematica I summed up the first 450 terms in each series. The result is 0.827760002939144239418158294687, which has an error of $10^{-23}$. I think this is the best we can extract out of this problem. I will stop working on it.
Best Answer
You mentioned that your main difficulty was finding $2\pi$ as a period. To do this, we can look at each term individually.
Let's look at the second integral first. All the terms look like $\sin(kx)$ or $\cos(kx)$, the period of which you may know to be $2\pi/k$. In particular, as long as $k$ is an integer, all of these are periodic with period $2\pi$ (that may not be the minimal period, say, for $\sin(1997x)$, but it is certainly a period). So, the integrand is periodic with period $2\pi$.
The first integral is a bit trickier, since the terms aren't all like $\sin(kx)$. However, note that $\sin\sin x$ should be periodic with period $2\pi$ (again, maybe there's some other smaller period, or maybe not). From this you should be able to convince yourself that a function like $\sin(\sin x+2016x)$ should be periodic with period $2\pi$; just plug in $x+2\pi$ and simplify.