Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$. 
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calculation.
Any ideas?
 A: *

*A behaviour as $n \to \infty$. Laplace's method ($3.$, p. $2$)
may be applied here, $$ \begin{align} \int_a^bf(x)e^{-\lambda
   g(x)}dx\sim f(a)e^{-\lambda g(a)}\sqrt{\frac{\pi}{2\lambda
   g''(a)}},\qquad \lambda \to \infty, \tag1 \end{align} $$ with
$g'(a)=0$, $g''(a)>0$, $f(a)\neq 0$. 
One may write $$ \begin{align} \int_{0}^{\large \frac \pi 2} \left( 1
 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
   \right )^{-\frac{n-1}{2}} d \theta \\\\&=\int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx
   \\\\&=\int_{0}^{\infty} \frac{1}{1+x^2}e^{-\large \frac{n-1}{2}\cdot
   \ln(1 + t^2 x^2)} dx \end{align} $$ here $ \lambda =\frac{n-1}{2}$,
$g(x)=\ln(1 + t^2 x^2)$, $g(0)=0$, $g'(0)=0$, $g''(0)=2t^2>0$,
$f(0)=1\neq 0$, we then get

$$ \int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\large\frac{n-1}{2}} d \theta \sim
   \frac1t\sqrt{\frac{\pi}{2}}\cdot
   \frac1{\sqrt{n}}=\mathcal{O}\left(\frac1{\sqrt{n}}\right),\qquad n
   \to \infty. \tag2 $$


*A closed form. One may evaluate $$ \int_{0}^{\large \frac \pi 2}
   \left( a +1+ \cot^2\theta \right )^{-\frac{1}{2}} d
   \theta=\frac{\arctan \sqrt{a}}{\sqrt{a}},\qquad a>0 $$ then
differentiating $n$ times with respect to $a$ setting
$t=\dfrac1{\sqrt{a+1}}$, gives

$$ \int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{-\frac{2n+1}{2}} d \theta=\frac{(-1)^n}{t^{2n+1}}
   \frac{(2n)!}{2^{2n-1}n!}\left[\frac{d^n}{da^n}\left(\frac{\arctan
   \sqrt{a}}{\sqrt{a}}\right)\right]_{t=\frac1{\sqrt{a+1}}}.\tag3 $$

Similarly, from the evaluation $$ \int_{0}^{\large \frac \pi 2}
   \left(a+1+\cot^2\theta\right)^{-1} d
   \theta=\dfrac1{a+1+\sqrt{a+1}},\qquad a>0 $$ then differentiating $n$
times with respect to $a$ setting $t=\dfrac1{\sqrt{a+1}}$ we get

$$ \int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{\large-\frac{2n+2}2} d \theta=\frac{(-1)^n}{t^{2n+2}}
   \frac{1}{n!}\left[\frac{d^n}{da^n}\left(\dfrac1{a+1+\sqrt{a+1}}\right)\right]_{t=\frac1{\sqrt{a+1}}}.
   \tag4 $$

A: One may write:
$$ 
\begin{align} 
I(n)=\int_{0}^{\large \frac \pi 2} \left( 1
 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
   \right )^{-\frac{n-1}{2}} d \theta \\\\
&=\int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx
\end{align}
$$
Upper bound ($t\leq1$):
$$
\begin{align}
I(n) &=\int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&\leq\int_{0}^{\infty}
   \frac{1}{1+t^2x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
   \left( 1 + t^2 x^2 \right )^{-\frac{n+1}{2}} dx \\\\
&= \frac1t \int_{0}^{\infty}
   \left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx = \frac1t f(n)
\end{align}
$$
Lower bound(for $t\leq 1$):
$$
\begin{align}
I(n) &\geq \int_{0}^{\infty}
   \frac{1}{1+x^2}\left( 1 + x^2 \right )^{-\frac{n-1}{2}} dx \\\\
&= \int_{0}^{\infty}
   \left( 1 + x^2 \right )^{-\frac{n+1}{2}} dx =f(n)\\\\
\end{align}
$$
Thus:
$$ f(n) \leq I(n) \leq \frac1t f(n)$$
which is valid as a finite approximation when $t$ is strictly larger than 0.
A recursion can be obtain for $f(n)$. Laplace approximation for $f(n)$ would give that $f(n) \approx \sqrt{\pi \over 2n}$ which show that the upper bound can be tight. 
