How to evaluate $\int_{\pi/2}^\pi \sqrt{1 - \frac{1}{2}\cos^2 x + \sin x \sin 2x} \;\mathrm{d}x$

Modification,

$$1-\frac{1}{2}\cos^2x +\sin x\sin 2x=\frac{(\sin x+\sin 2x)^2+(\cos 2x+2)^2}{2}$$

Suppose :

$$\frac{\sin x+\sin 2x}{\cos 2x+2}= \tan \alpha$$

$$\alpha= \tan^{-1}\frac{\sin x+\sin 2x}{\cos 2x+2}$$

Then we have:

$$I=\int1-\frac{1}{2}\cos^2x +\sin x\sin 2x=\frac{1}{2}\int \frac{(\cos 2x+2)}{\cos \alpha} dx$$

Suppose we can write:

$$I=\frac{1}{2}\int^{\pi}_{\pi/2}(\cos 2x +2)dx\int^a_b \frac{d \alpha}{\cos \alpha}$$

If we find a and b as functions of x it will give the initial form of integrand, to avoid this let's assume we can calculate a and b for interval $[\pi/2, \pi]$ as follows:

$a=\tan^{-1}\frac{\sin x+\sin 2x}{\cos 2x +2}$ for $x=\pi/2$, which gives $a=\pi/4$ and for $x=\pi$ which gives $b=0$, then we have:

$$I_1=\int^{\pi/4}_0 \frac{d \alpha}{\cos \alpha}=\frac{1}{2}[\ln \frac{1+\tan \frac{\alpha}{2}}{1-\tan \frac{\alpha}{2}}]^{\pi/4}_0≈0.44$$

Finally we get:

$$I=\frac{1}{2}\times 0.44 \int^{\pi}_{\pi/2}(\cos 2x +2)dx=0.22\times [\frac{1}{2}\sin 2x + 2x]^{\pi}_{\pi/2}≈ 0.7$$

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.