I need to calculate the following definite integral:
\begin{align} I = \int^{\pi}_{0}\frac{\pi^2-x^2}{\sqrt{1-\sqrt{1-a\cos^{2}\frac{x}{2}}}} dx \end{align} where $a$ is a real number between $0$ and $1$.
Is there an analytic expression for this integral (for example, in terms of some special functions of a)? Wolfram Mathematica failed to give me any answer.
I am afraid that, beside the cases where $a=0$ or $a=1$, there is no explicit solution for this definite integral; the problem is that $0 < a <1$ !
However, if $$I(a)=\int_0^\pi\frac{\pi ^2-x^2}{\sqrt{1-\sqrt{1-a \cos ^2\left(\frac{x}{2}\right)}}}\, dx$$ it seems that a reasonable approximation could be given by something like $$I(a)=\frac {\alpha - \beta\, a}{\sqrt a}$$
A linear regression gives, with $R^2=0.999994$, $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 47.6698 & 0.0146 & \{47.6407,47.6988\} \\ \beta & 3.72710 & 0.0469 & \{3.63400,3.82019\} \\ \end{array}$$
For $a=1$, this would give $I(1)\sim 43.9427$ while the exact value is $I(1)=43.1195$ which is the numerical value of $$\frac{1}{8} \left(896 \sqrt{2} \zeta (3)+8 \sqrt{2} \pi ^2 \left(\log \left(3-2 \sqrt{2}\right)-3 i \pi \right)+(1-i) \psi ^{(2)}\left(\frac{1}{8}\right)-(1+i) \psi ^{(2)}\left(\frac{3}{8}\right)-(1-i) \psi ^{(2)}\left(\frac{5}{8}\right)+(1+i) \psi ^{(2)}\left(\frac{7}{8}\right)\right)$$
The approximation is very good for small values of $a$. So, using $a=b^2$ and expanding as a series built around $b=0$, we have $$\frac{1}{\sqrt{1-\sqrt{1-b^2 \cos ^2\left(\frac{x}{2}\right)}}}=\sum_{n=0}^\infty F_n(x)\, b^{2n-1}$$ The first of these functions is $$F_0(x)= \frac{2}{\sqrt{\cos (x)+1}}$$ For $1 \leq n \leq 8$ they are odd powers of $\cos \left(\frac{x}{2}\right)$ and, for $n \geq 9$ they are even powers of $\cos \left(\frac{x}{2}\right)$ divided by $\sqrt{1+\cos (x)}$.
The computation of the definite integrals $$J_n=\int_ 0^\pi (\pi^2-x^2)\cos ^{2 n+1}\left(\frac{x}{2}\right)\,dx \quad \text{and}\quad K_n=\int_ 0^\pi (\pi^2-x^2)\frac{\cos ^{2 n}\left(\frac{x}{2}\right)}{\sqrt{\cos (x)+1}}\,dx$$ do not present any problem (thanks to Wolfram Alpha) and the final result write $$I(a)=\frac{28 \sqrt{2} \zeta (3)}{b}-\frac 1 {\sqrt 2}\sum_{n=0}^\infty \frac {\alpha_n}{\beta_n} b^{2n+1}$$ The first coefficients are given below $$\left( \begin{array}{ccc} n & {\alpha_n} & {\beta_n}\\ 0 & 4 & 1 \\ 1 & 25 & 27 \\ 2 & 1813 & 4500 \\ 3 & 461747 & 2058000 \\ 4 & 285567139 & 2000376000 \\ 5 & 30050088913 & 304285766400 \\ 6 & 18059147415499 & 249579242744832 \\ 7 & 13141453098161 & 237694516899840 \\ 8 & 55563794154986291 & 1273956176213360640 \\ 9 & 19575535815512966389 & 554797803977615278080 \\ 10 & 1608205297319647165903 & 55206649171187932594176 \\ 11 & 1991247300342258043248715 & 81418097026162857681616896 \\ 12 & 9371208288978433635255412703 & 450014992066376339226624000000 \end{array} \right)$$
For $b=1$, the above gives $$28 \sqrt{2} \zeta (3)-\frac{33988692851436059475103482406057}{5613487011035978455512907776000 \sqrt{2}}=43.3176\cdots$$ still too high but, taking more terms, it should converge.