How to evaluate $\int_{\pi/2}^\pi \sqrt{1 - \frac{1}{2}\cos^2 x + \sin x \sin 2x} \;\mathrm{d}x$ Here is an integral from a very old math journal:
(https://www.jstor.org/stable/1967417)
$$\int_{\pi/2}^\pi \sqrt{1 - \frac{1}{2}\cos^2 x + \sin x \sin 2x} \;\mathrm{d}x$$
It appears the journal never published a solution to this problem.
My attempts:


*

*If we make the substitution $t = \cos x$, then the integral equals
$$\int_{-1}^0 \sqrt{\frac{1 - \frac{t^2}{2} + 2t(1 - t^2)}{1 - t^2}} \;\mathrm{d}t$$
However it appears impossible to integrate this function. Mathematica fails to evaluate both the indefinite and the definite integral.

*If we integrate this function numerically, we get 0.827760002939144239418157727592. The Inverse Symbolic Calculator returns nothing for this number. WolframAlpha gives nothing interesting either.
Update: The integral can be expressed as a double sum. However it converges slowly. I am now looking for a faster way to compute its value.
 A: 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:
$$I=\frac{1}{2}\times 0.44 \int^{\pi}_{\pi/2}(\cos 2x +2)dx=0.22\times [\frac{1}{2}\sin 2x + 2x]^{\pi}_{\pi/2}≈ 0.7$$
