How to approach an integral over $g(\cos(t))$ from $0$ to $2\pi$, where $g(x)$ is nasty?

Question

For notational convenience, let $f(t) = a^2 + 2 a b \cdot \cos(t) + b^2$, where $a,b$ are both positive real constants and $t$ will be the variable of integration, which is supposed to be carried out from $t=0$ to $t=2 \pi$. I want to find an expression (or approximations) for \begin{align} \int\limits_0^{2\pi} \frac{\exp(-c(24m^2 - 24m \sqrt{f(t)} + 7 f(t) - 4 a b \cdot \cos(t)))}{(m^2\cdot f(t))^{1/4}} dt \end{align} where $c,m$ are again positive real constants.

My first attempt was to use Mathematica, but without any result. Are there maybe any tricks how to approximate the above integral? Any comments or suggestions are most welcome! Thanks in Advance.

Answer 1

Because of symmetry, we will take the integral from $t=0$ to $t=\pi$ and double it. Let $f(t)=u$.

It follows that $$du=-2abSIN(t)dt$$ Converting the limits ($t=0$ & $t=\pi$) from t to u using the equation $u=a^2+2abCOS(t)+b^2$ we get $$lower limit=a^2+2ab+b^2$$ $$upper limit=a^2-2ab+b^2$$ Solving for cos(t), we get $$cos(t)=\frac{u-a^2-b^2}{2ab}$$

$$t=ARCCOS\left(\frac{u-a^2-b^2}{2ab}\right)$$ Substituting these into the integral, we get $$\int\limits_0^{\pi} \frac{e^{-c\left[24m^2 - 24m \sqrt{f(t)} + 7 f(t) - 4 a b \cdot \cos(t)\right]}}{[m^2 f(t)]^{1/4}} dt=\int_{a^2+2ab+b^2}^{a^2-2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+7u-4ab\left(\frac{u-a^2-b^2}{2ab}\right)\right]}}{-2abSIN(t)(m^2u)^{\frac{1}{4}}}du$$ Because we are integrating from $0$ to $\pi$ we can use $$sin(t)=\sqrt{1-cos^2(t)}$$ We get $$\int_{a^2+2ab+b^2}^{a^2-2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+7u-4ab\left(\frac{u-a^2-b^2}{2ab}\right)\right]}}{-2ab\sqrt{1-\left(\frac{u-a^2-b^2}{2ab}\right)^2}m^{\frac 12}u^{\frac{1}{4}}}du$$ $$=\int_{a^2-2ab+b^2}^{a^2+2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+5u+2(a^2+b^2)\right]}}{2ab\sqrt{1-\left(\frac{u-a^2-b^2}{2ab}\right)^2} m^{\frac 12}u^{\frac{1}{4}}}du$$