An integral and a mysterious recursive sequence Today I tried to solve the integral
$$I_n= \int_0^{\frac{\pi}{2}} \frac{\mathrm{dt}}{(a \cos^2(t)+ b \sin^2(t))^n} \quad \quad \quad \quad a,b > 0$$
After some dirty calculations, I obtained that
$$I_n = \frac{\pi}{2^n(n-1)!\sqrt{ab}} \sum_{k=0}^{n-1} \frac{C(n-1,k)}{a^kb^{n-1-k}}$$
where the $C(n,k)$ are defined for $n \geq 1$ and $0 \leq k \leq n$ and satisfy the "Pascal-like" recursive relation
$$C(n,k)=(2k-1) \times C(n-1,k-1)+(2n-2k-1) \times C(n-1,k) \quad \quad n\geq 2, \text{ } k \geq 1$$
with the initial conditions $$C(1,0)=C(1,1)=1 \quad \quad \text{and} \quad \quad C(n,0)=C(n,n)=\frac{(2n)!}{2^nn!} \text{ for all } n$$
My question are :

*

*is that a "well-known" sequence of coefficients ? Do you know if they appear in some other problems ?

*are also these integrals well-known ? Do you maybe know an elegant way to solve them ?

In a summary, can you tell me anything you know about these integrals and this recursive sequence !
Thanks a lot !
EDIT : As @ReneGy pointed out, this sequence is apparently registered is OEIS as A059366. But OEIS does not mention the recursive relation I gave... I am still taking anything you can tell me about this mysterious sequence and these integrals !
 A: Now, the new triangular table for $C(n,k)$




n\k
1
2
3
4
5
6
7




1
1
1







2
3
2
3






3
15
9
9
15





4
105
60
54
60
105




5
945
525
450
450
525
945



6
10395
5670
4725
4500
4725
5670
10395




This is A059366 in the OEIS.
Apparently, these numbers were already found for the calculation of $$I_n= \int_0^{\frac{\pi}{2}} \frac{\mathrm{dt}}{(a \cos^2(t)+ b \sin^2(t))^n} \quad \quad \quad \quad a,b > 0,$$see   L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.
We have:
$$  \sum_{n\ge 0}I_nx^n=\int_0^{\frac{\pi}{2}}\mathrm{dt}\sum_{n\ge 0} \frac{x^n}{(a \cos^2(t)+ b \sin^2(t))^n}= \int_0^{\frac{\pi}{2}}\mathrm{dt}\frac{1}{1-\frac{x}{a \cos^2(t)+ b \sin^2(t)}}$$
$$  \sum_{n\ge 0}I_nx^n= \int_0^{\frac{\pi}{2}}\frac{a \cos^2(t)+ b \sin^2(t)}{a \cos^2(t)+ b \sin^2(t)-x}\mathrm{dt}=\frac{\pi}{2}+\int_0^{\frac{\pi}{2}}\frac{x}{a \cos^2(t)+ b \sin^2(t)-x}\mathrm{dt}$$
$$  \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}+\int_0^{\frac{\pi}{2}}\frac{x}{a-x + (b-x) \tan^2(t)}\mathrm{d\tan{t}}=\frac{\pi}{2}+\int_0^{\infty}\frac{x}{a-x + (b-x) t^2}\mathrm{dt}$$
$$  \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}+\frac{x}{a-x}\int_0^{\infty}\frac{1}{1 +  (\frac{\sqrt{b-x}}{\sqrt{a-x}}t)^2}\mathrm{dt}=\frac{\pi}{2}+\frac{x}{\sqrt{(a-x)(b-x)}}\int_0^{\infty}\frac{1}{1 +  t^2}\mathrm{dt}$$
$$  \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}\big(1+\frac{x}{\sqrt{(a-x)(b-x)}}\big)$$
then , for $n\gt 0$
$$I_n = \frac{\pi}{2} [x^{n}]\frac{x}{\sqrt{(a-x)(b-x)}}$$
but $$  \sum_{i\ge 0}{2i\choose i}\big(\frac{x}{a}\big)^i=\frac{\sqrt a}{\sqrt{a-4x}}.$$
Then we have
$$I_n = \frac{\pi}{2^{2n-1}\sqrt{ab}} \sum_{i+j=n-1} \frac{{2i\choose i}}{a^i}\frac{{2j\choose j}}{b^j}.$$
By identification to the OP formula, this gives
$$C(n,k)=  \frac{n!}{2^n}{2k\choose k}{2(n-k)\choose n-k} $$
which is
$$C(n,k)=  (-2)^n{n\choose k}{-\frac{1}{2}\choose k} {-\frac{1}{2}\choose n-k}$$ which is given in the above OEIS link (with a typo).
It should be not so difficult to verify that these closed formulae for $C(n,k)$ do satisfy the above recursion.
A: Note that
\begin{align}
& \left( \frac{(b-a)\sin t\cos t}{(a \cos^2 t+ b \sin^2 t)^{n-1}} \right)’\\
= & -\frac{{2(n-1)ab}}{(a \cos^2 t+ b \sin^2 t)^n}
 + \frac{{(2n-3)(a+b)}}{(a \cos^2 t+ b \sin^2 t)^{n-1}}
- \frac{{2(n-2)}}{(a \cos^2 t+ b \sin^2 t)^{n-2}}
\end{align}
Integrate both sides to establish
$$(n-1)I_n =  (n-\frac32)\left(\frac1a+\frac1b\right)I_{n-1}-\frac{n-2}{ab}I_{n-2}
$$
with $I_1 =\frac\pi{2\sqrt{ab}}$. Then
\begin{align}
& I_2 = \frac12 I_1\left(\frac1a+\frac1b\right) \\
& I_3 = \frac18I_1\left(\frac3{a^2}+\frac2{ab}+\frac3{b^2}\right)\\
&\>\cdots\\
& I_n = I_1
\left(\frac{c_{n,0}}{a^{n-1} }+ \frac{c_{n,1}}{a^{n-2}b} + \cdots + \frac{c_{n,n-1}}{b^{n-1} }\right)
\end{align}
where the coefficients satisfy the recursive relationship $c_{n,0}=c_{n,n-1}=\frac{(2n-3)!!}{(2n-2)!!}$
$$c_{n,k} = \frac{2n-3}{2(n-1)}(c_{n-1,k} +c_{n-1,k-1})
 - \frac{n-2}{n-1}c_{n-2,k-1}$$
A: There is an antiderivative for
$$J_n= \int \frac{\mathrm{dt}}{\big[a \cos^2(t)+ b \sin^2(t)\big]^n} $$
(have a look here) in terms of the Appell hypergeometric function of two variables.
Using the integration bounds,
$$I_n=-\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}} \,\Gamma (2-n) }{2 (n-1) (a-b) (ab)^n \,\Gamma \left(\frac{3}{2}-n\right)} \,A$$ where
$$A=a^n b\,\, _2F_1\left(\frac{1}{2},1-n;\frac{3-2n}{2};\frac{b}{a}\right)+a b^n \,\,
   _2F_1\left(\frac{1}{2},1-n;\frac{3-2n}{2};\frac{a}{b}\right)$$ where appears the gaussian hypergeometric function.
A: Start with
$$J_n(p)=\frac2\pi \int_0^{\frac{\pi}{2}} \frac{{dt}}{(p \cos^2t+ \sin^2t)^n}$$
$$J_{n+1}(p)=J_n(p)+\frac{p-1}n J_n’(p)\tag 1
$$
With $J_1(p)= \frac1{p^{1/2}} $
\begin{align}
& J_2 (p)=\frac12\left(\frac1{p^{1/2}}+\frac1{p^{3/2}}\right)\\
& J_3 (p)= \frac18\left(\frac3{p^{1/2}}+\frac2{p^{3/2}}+\frac3{p^{5/2}}\right)\\
& J_4 (p)= \frac1{16}\left(\frac5{p^{1/2}}+\frac3{p^{3/2}}+\frac3{p^{5/2}}+\frac5{p^{7/2}}\right)\\
& \cdots\>\cdots\\
& J_{n}(p) = \frac{c_{n-1,0} }{p^{1/2}}+ \frac{c_{n-1,1} }{p^{3/2}}+\cdots + \frac{c_{n-1,n-1} }{p^{(2n-1)/2}}\\
& J_{n+1}(p) = \frac{c_{n,0} }{p^{1/2}}+ \frac{c_{n,1} }{p^{3/2}}+\cdots + \frac{c_{n,n-1} }{p^{(2n-1)/2}}+\frac{c_{n,n} }{p^{(2n+1)/2}}
\end{align}
with the coefficients satisfying the recursion $c_{0,0}=1,\>c_{n,0}= \frac{2n-1}{2n}c_{n-1,0}$
$$c_{n,k}=\frac{k-n-1}kc_{n,k-1} +\frac nk c_{n-1,k-1}, \>\>\>\>\>k\ne 0$$
Then
\begin{align}
I_{n+1} &= \int_0^{\frac{\pi}{2}} \frac{{dt}}{(a \cos^2t+ b\sin^2t)^{n+1}}
=\frac\pi2 \frac1{b^{n+1}} J_{n+1}\left(\frac ab \right)\\
&=  \frac\pi2\left(\frac{c_{n,0} }{a^{1/2}b^{(2n+1)/2}}+ \frac{c_{n,1} }{a^{3/2}b^{(2n-1)/2}}+\cdots + \frac{c_{n,n} }{a^{(2n+1)/2}b^{1/2}}\right)
\end{align}
A: Not an answer, but too long for comment: here is a triangular table for $C(n,k)$




n\k
0
1
2
3
4
5
6




0
1








1
0
1







2
0
1
3






3
0
3
6
15





4
0
15
27
45
105




5
0
105
180
270
420
945



6
0
945
1575
2250
3150
4725
10395




It does not seem to be in the OEIS.
