Here it is:
$$\int_{K_1}^{K_2}\Big(\frac{\pi}{2}+\arcsin(\cos\varphi+0.75)+\arcsin(\sin\varphi+0.75)\Big)\,\,\mathrm{d}\varphi$$
Where $K_1=\pi+\arccos\big(\frac{3}{4}\big)$ and $K_2=\frac{3\pi}{2}-\arccos\big(\frac{3}{4}\big)$. This is coming from a circle-type question I made. The main problem is the $\arcsin(\sin\varphi+0.75)$. WolframAlpha can't simplify this value alone, and I'm starting to believe this can't be simplified. However, maybe the integral (along with the $\arcsin(\cos \varphi+0.75)$, which may combine with other arcsine and simplify) can, and I was wondering whether the general form $\arcsin(\sin\varphi+C)$ can be simplified at all.
I attempted to find the value of the arcsine alone, so I drew a triangle; and adding a constant to arcsine's argument is essentially increasing a side length of a right triangle, which was actually way harder than I expected.
Similarly, I gave up on substitution (I removed the bounds of integration for now):
\begin{align*}\int\arcsin(\sin\varphi+0.75)\;\mathrm{d}\varphi&=\int\frac{\arcsin u \;\mathrm{d}u}{\sqrt{1-\big(u-\frac{3}{4}\big)^2}}\;,\;\varphi=\arcsin(u-0.75)\\&\stackrel{IBP}{=}-\arcsin\bigg(\frac{3}{4}-u\bigg)\arcsin u+\int\frac{\arcsin\big(\frac{3}{4}-u\big)\;\mathrm{d}u}{\sqrt{1-u^2}}\end{align*}
I thought about bringing the integral over the equal sign, to get something like $2\int a=b$, and I would get the integral's value. However, I couldn't just do that because $u$ wasn't a dummy variable under the sign.
Question: I am wondering if my solution was correct so far, and I am also looking for a complete solution to the problem.
Best Answer
Why not to try a series solution around $x=0$ ?
$$I=\int_{K_1}^{K_2}\Bigg(\frac{\pi }{2}+\sin ^{-1}\left(\cos (x)+\frac{3}{4}\right)+\sin ^{-1}\left(\sin (x)+\frac{3}{4}\right)\Bigg)\,dx$$ $$I=2\int_{K_1}^{\frac {5\pi}4}\Bigg(\frac{\pi }{2}+\sin ^{-1}\left(\cos (x)+\frac{3}{4}\right)+\sin ^{-1}\left(\sin (x)+\frac{3}{4}\right)\Bigg)\,dx$$
Expanded as series around $x=\frac {5\pi}4$, the integrand write $$\frac{1}{2} \left(\pi +4 \sin ^{-1}\left(\frac{3}{4}-\frac{1}{\sqrt{2}}\right)\right)+\sum_{n=1}^\infty a_n\,\left(x-\frac{5 \pi }{4}\right)^{2n}$$ and the first coefficients $a_n$ are $$\left( \begin{array}{cc} n & a_n \\ 1 & \frac{18}{287} \sqrt{\frac{2}{287} \left(27968 \sqrt{2}-20151\right)} \\ 2 & -\frac{3 \left(5471 \sqrt{2}-9100\right)}{2 \left(12 \sqrt{2}-1\right)^{7/2}} \\ 3 & \frac{148855716-104047681 \sqrt{2}}{20 \left(12 \sqrt{2}-1\right)^{11/2}} \\ 4 & \frac{857639817332-603034132649 \sqrt{2}}{160 \left(12 \sqrt{2}-1\right)^{15/2}} \\ 5 & \frac{433099352425458692-305868400361017345 \sqrt{2}}{100800 \left(12 \sqrt{2}-1\right)^{19/2}} \\ 6 & \frac{49027241222149014535884-34660027738879924897247 \sqrt{2}}{13305600 \left(12 \sqrt{2}-1\right)^{23/2}} \\ 7 & \frac{1143818651275076309270210428-808763022809697073020837367 \sqrt{2}}{345945600 \left(12 \sqrt{2}-1\right)^{27/2}} \\ 8 & \frac{1781445227570529438648811322348140-125965894251168683614 9931462848799 \sqrt{2}}{581188608000 \left(12 \sqrt{2}-1\right)^{31/2}} \\ \end{array} \right)$$
Using
$$\int_{K_1}^{K_2}\left(x-\frac{5 \pi }{4}\right)^{2 n}\,dx=\frac{ \left(\pi -4 \cos ^{-1}\left(\frac{3}{4}\right)\right)^{2 n+1}}{(2 n+1)\,\,2^{4 n+1}}$$
computing the partial sums and converting the results to decimal values
$$\left( \begin{array}{cc} p & \sum_{n=1}^p \\ 1 & 0.2077388562120074416639450 \\ 2 & 0.2077389047478118467180908 \\ 3 & 0.2077389047701361779155620 \\ 4 & 0.2077389047702303755243590 \\ 5 & 0.2077389047702305849757772 \\ 6 & 0.2077389047702305853829307 \\ 7 & 0.2077389047702305853840193 \\ 8 & 0.2077389047702305853840222 \\ \cdots & \cdots \\ \infty &0.2077389047702305853840223\\ \end{array} \right)$$