In one of my problems, I encountered the integral: $$\int_0^\pi \sqrt{1+\cos(mx)} \cos(\rho \sin(x)) dx$$ where $\rho$ and $m$ are parameters. I would be satisfied if the answer was given for $m=1$, and if the square root would be expanded using the Taylor series.
Naturally, I have already tried all of this and checked it in Wolfram Alpha. But it said there was no answer. I am also definitely not interested in numeric integration
As you wrote, the idea is to develop the square root using the binomial them.
So, for $m=1$, we face integrals $$I_n=\int_0^\pi \cos ^n(x) \cos (\rho \sin (x))\,dx$$
Using a CAS $$I_n=\sqrt{\pi }\, 2^{\frac{n}{2}-1} \left(1+(-1)^n\right) \rho ^{-n/2} \,\Gamma \left(\frac{n+1}{2}\right) J_{\frac{n}{2}}(\rho )$$ provided $\rho \in \mathbb{R}\land \Re(n)>-1$.
This means $$\color{blue}{I_{2n}=\sqrt{\pi }\, \left(\frac 2 \rho\right)^n\, \Gamma \left(n+\frac{1}{2}\right) J_n(\rho )}$$ and $I_{2n+1}=0$.
So, for $$K=\int_0^\pi \sqrt{1+\cos(x)} \cos(\rho \sin(x))\, dx$$
$$\color{red}{K=\sqrt{\pi }\,\sum _{n=0}^{\infty } \binom{\frac{1}{2}}{2 n} \, \left(\frac 2 \rho\right)^n\, \Gamma \left(\frac{2n+1}{2}\right) J_n(\rho )}$$ which will probably converge very slowly since, for large values of $n$ ($a_n$ being the summand), for any $\rho$ $$\frac{a_{n+1}}{a_n}=\frac{\left(16 n^2-1\right) J_{n+1}(\rho )}{8 (n+1)\, \rho \, J_n(\rho )}=1-\frac{2}{n}+\frac{137}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ Computing the sums up to $p$ $$\left( \begin{array}{cccc} p & \rho=\frac 12& \rho= 1& \rho= 2 \\ 100 & 2.644596475 & 2.122507198 & 0.4988469627 \\ 200 & 2.643719191 & 2.121631131 & 0.4979757517 \\ 300 & 2.643425787 & 2.121337955 & 0.4976834850 \\ 400 & 2.643278902 & 2.121191150 & 0.4975369993 \\ 500 & 2.643190713 & 2.121102998 & 0.4974489948 \\ 600 & 2.643131895 & 2.121044200 & 0.4973902779 \\ 700 & 2.643089870 & 2.121002188 & 0.4973483140 \\ 800 & 2.643058345 & 2.120970671 & 0.4973168285 \\ 900 & 2.643033822 & 2.120946153 & 0.4972923322 \\ 1000 & 2.643014201 & 2.120926535 & 0.4972727305 \\ \cdots & \cdots & \cdots & \cdots \\ \infty &2.642837512 &2.120749860 & 0.4970961248 \end{array} \right)$$
Edit
Looking at the numbers given in the table, we could notice that the difference between the summations up to $100$ and up to $\infty$ are almost constants (respectively $0.00175896$, $0.00175734$ and $0.00175084$).
So, we could approximate the infinite summation writing $$K=\sqrt{\pi }\,\sum _{n=0}^{p } \binom{\frac{1}{2}}{2 n} \, \left(\frac 2 \rho\right)^n\, \Gamma \left(\frac{2n+1}{2}\right) J_n(\rho )+$$ $$\sum _{n=p+1}^{\infty }\sqrt{\pi }\, \binom{\frac{1}{2}}{2 n} \, \left(\frac 2 \rho\right)^n\, \Gamma \left(\frac{2n+1}{2}\right) J_n(\rho )$$ and, as already done for the ratio test, use for large orders $$J_n(\rho) \sim \frac{1}{\sqrt{2 \pi n} }\left(\frac{e\rho}{2n}\right)^n $$ which makes for the second summation $$\sum _{n=p+1}^{\infty }\sim \frac{1}{\sqrt{2}}\sum _{n=p+1}^{\infty }e^n n^{-(n+\frac{1}{2})} \binom{\frac{1}{2}}{2 n} \Gamma \left(n+\frac{1}{2}\right)$$
Using composition of Taylor expansions to $O\left(\frac{1}{n}\right)$ we have $$\sum _{n=p+1}^{\infty }\sqrt{\pi }\, \binom{\frac{1}{2}}{2 n} \, \left(\frac 2 \rho\right)^n\, \Gamma \left(\frac{2n+1}{2}\right) J_n(\rho )\sim \frac{\psi ^{(1)}(p+1)}{4 \sqrt{2}}$$ which, for $p=100$ gives $\approx 0.00175896$ (!!)