Consider the integral: $$\int_{0}^{\pi}\frac{dx}{\pi}\frac{\cos(xt)}{\sqrt{1+a\sin^{2}(x/2)}},\,\,\,\,\,\,\mathrm{as\,\,}t\rightarrow\infty$$
with $a>0.$
I know that the answer is $\exp(-t/\xi)$, where $\xi=1/\ln(2)=1.4427$ for $a=8$
I'm interested in the leading behavior of this integral as $t\rightarrow\infty$ . I tried to applied the Laplace's Method but did not get the answer.
Please help me.
Let $f(t)$ be represented by the integral
$$f(t)=\frac1\pi \int_0^\pi \frac{\cos(xt)}{\sqrt{1+\sin^2(x/2)}}\,dx \tag 1$$
Integrating by parts the integral in $(1)$ with $u=\frac{1}{\sqrt{1+\sin^2(x/2)}}$ and $v=\frac{\sin(xt)}{t}$ reveals
$$\begin{align} f(t)&= \frac{\sin(\pi t)}{\pi \sqrt{2}\,t}+\frac1{4\pi t} \int_0^\pi \frac{\sin(xt)\sin(x)}{\left(1+\sin^2(x/2)\right)^{3/2}}\,dx \tag 2\\\\ &=\frac{\sin(\pi t)}{\pi \sqrt{2}\,t}+o\left(\frac1{t^2}\right) \tag3 \end{align}$$
Note that integrating by parts the integral in $(2)$ with $u=\frac{\sin(x)}{\left(1+\sin^2(x/2)\right)^{3/2}}$ and $v=-\frac{\cos(xt)}{t}$ would produce a term that is of order $o\left(\frac{1}{t^2}\right)$, thereby producing $(3)$.