I am trying to solve: $$\int_a^b\frac{dx}{\sqrt{1-(A-\cos(Bx))^2}}.$$ I have first tried to use the substitution $u=\tan(x/2)$ then after simplifying $v=\sqrt{\frac{A}{A+2}}\tan(u)$, this gives an elliptic function, but then when I evaluate with $a=0$ and $b=\frac{1}{B}\arccos(A-1)$ it gives an indeterminate answer.
Any suggestions would be greatly appreciated!
Let $Bx=t$ to make the integral $$J= B\,I=\int_0^{k \cos ^{-1}(A-1)}\frac{dt}{\sqrt{1-(A-\cos (t))^2}}$$ where $0\leq A\leq 2$ and where "hope" to make $k \to 1$.
For the antiderivative, we face the elliptic integral which for $t=0$ gives $0$. Then, the only problem is at the upper bound where we seem to face a serious problem of the type $\frac 00$ when $k \to 1$.
Using L'Hospital rule six times did not solve the problem.
Looking in more details at the series expansions of both numerator and denominator show that they both are polynomials in $\sqrt{k-1}$ without constant term. However, for the numerator, the first term remains undefined !
Then, I used brute force and showed that for $k=1-\epsilon$, the calculations work nicely.
Working with illimited precision, I have been able to use $\epsilon=10^{-48}$ and compared with numerical integration with $k=1$. $$\left( \begin{array}{ccc} A & \text{using } k=1-10^{-48} & \text{numerical integration} \\ 0.1 & 4.384143232 & 4.384143271 \\ 0.2 & 3.695637363 & 3.695636872 \\ 0.3 & 3.296411566 & 3.296411602 \\ 0.4 & 3.016112492 & 3.016112507 \\ 0.5 & 2.801206085 & 2.801206087 \\ 0.6 & 2.627773332 & 2.627773333 \\ 0.7 & 2.483014115 & 2.483014115 \\ 0.8 & 2.359263555 & 2.359263555 \\ 0.9 & 2.251562589 & 2.251562589 \\ 1.0 & 2.156515647 & 2.156515648 \\ 1.1 & 2.071694646 & 2.071694646 \\ 1.2 & 1.995302778 & 1.995302778 \\ 1.3 & 1.925972938 & 1.925972938 \\ 1.4 & 1.862640802 & 1.862640802 \\ 1.5 & 1.804461622 & 1.804461622 \\ 1.6 & 1.750753803 & 1.750753803 \\ 1.7 & 1.700959543 & 1.700959543 \\ 1.8 & 1.654616668 & 1.654616668 \\ 1.9 & 1.611338059 & 1.611338059 \end{array} \right)$$