Asymptotic solution to the integral $\int_{-\pi/2}^{\pi/2} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x$ Recently, I have posted a question on how to find a reduction formula for the trigonometric integral
$$\int (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x.$$
This problem seems to be tough, however. When trying to solve (as proposed in a comment on my previous question) the integral using the Binomial theorem, one obtains a sum that seems to be quite complicated.
Thus, there seems to be only a little chance for a reasonable reduction formula to exist. As a consequence, I became interested in solving the integral asymptotically. So my question is how to obtain a reasonable asymptotic estimate for the following definite integral, as $n \to \infty$:
$$\int_{-\pi/2}^{\pi/2} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x.$$
I have tried to use the Laplace's method, but with my small knowledge and without any experience, I have not succeeded. Thus, I would be grateful for hints.
 A: Hint: Use the substitution $$u=\alpha +\sin{x},$$ $$du=\cos{x}\space dx,$$ $$\cos{x}=\sqrt{1-\sin^2{x}}=\sqrt{1-(u-\alpha)^2}.$$
The third equation is valid on intervals of $x$ where cosine is nonnegative, as is the case for $-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}.$ Then,
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x=\int_{\alpha -1}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}du.$$
Applying a simple shift substitution, $v=u-\alpha$, the integral becomes
$$ \int_{\alpha -1}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}du = \int_{-1}^{1} (\alpha+v)^n\sqrt{1-v^2}dv .$$
A: Let me know if any of this needs more details but there is a a fair bit of algebra and it will cloud the discussion if I include it all. By symmetry and trig identities:
$$\int _{-\pi/2 }^{\pi/2 }\! \left( \alpha+\sin \left( x \right) 
 \right) ^{n} \left( \cos \left( x \right)  \right) ^{2}{dx}=\frac{1}{4}
\int _{-\pi }^{\pi }\! \left( \alpha+\sin \left( x \right)  \right) ^{
n}\cos \left( 2\,x \right) {dx}+\frac{1}{4}\int _{-\pi }^{\pi }\! \left( 
\alpha+\sin \left( x \right)  \right) ^{n}{dx} \tag{1}$$
and by two rounds of binomial expansion:
$$\left( \alpha+\sin \left( x \right)  \right) ^{n}=\sum _{k=0}^{n}
 {\frac {1}{{2}^{
k}}}{n\choose k}{\alpha}^{n-k}{i}^{k}\sum 
_{j=0}^{k}{k\choose j} \left( -1 \right) ^{j}{{\rm e}^{ix(2j-k)}} $$
so $(1)$ is equivalent to:
$$  \sum _{k=0}^{n}  \mathfrak{R}\left( {\frac {{n\choose k}{\alpha}^{n-
k}{i}^{k}\sum _{j=0}^{k} \left( {k\choose j} \left( -1 \right) ^{j}
\int _{-\pi }^{\pi }\!{{\rm e}^{ix \left( 2\,j+2-k \right) }}{dx}
 \right) }{{2}^{k+2}}} \right) +\sum _{k=0}^{n} \frac{ 1}{{2}^{k+2}}\left({n\choose k}{\alpha}^{n-k}{i}^{k}\sum _{j=0}^{k} {k
\choose j} \left( -1 \right) ^{j}\int _{-\pi }^{\pi }\!{{\rm e}^{ix
 \left( 2\,j-k \right) }}{dx} \right) \tag{2}$$ 
where the real part crops up because I am using only one exponential from the cosine. Then from the integral rep of the Kronecker delta symbol:
$$\int _{-\pi }^{\pi }\!{{\rm e}^{ix
 \left( 2\,j-k \right) }}{dx}=2\pi \delta_{0,2j-k}\quad,\quad \int _{-\pi }^{\pi }\!{{\rm e}^{ix
 \left( 2\,j+2-k \right) }}{dx}=2\pi \delta_{0,2j+2-k}$$
it is evident that we sum only over even $k$ picking out one $j$ term from each inner sum. After a bit of algebra and converting binomial coefficients to factorials we obtain:
$$\begin{aligned}
\int _{-1/2\,\pi }^{1/2\,\pi }\! \left( \alpha+\sin \left( x \right) 
 \right) ^{n} \left( \cos \left( x \right)  \right) ^{2}{dx}&=\frac{\pi}{2^{n+1}}\sum _{k=0}^{\lfloor  n/2 \rfloor }{
\frac {n!\, \left( 2\,\alpha \right) ^{n-2\,k}}{ \left( k+1 \right) 
 \left( k! \right) ^{2} \left( n-2\,k \right) !}}\\
&=\frac{\pi \,{
\alpha}^{n}}{2}
{\mbox{$_2$F$_1$}(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};\,2;\,\frac{1}{\alpha^2})}
 \tag{3} \end{aligned}$$
where $\mbox{$_2$F$_1$}$ is a hypergeometric function . Alternatively the result can also be given as a Jacobi polynomial but the order is integer for odd $n$ only:
$$\frac{\pi \,{
\alpha}^{n}}{2}
{\mbox{$_2$F$_1$}(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};\,2;\,\frac{1}{\alpha^2})}=\frac{{\alpha}^{n}\pi}{n+1} P_{\frac{1}{2}(n-1)}^{1,-3/2-n}\left(1-\frac{2}{\alpha^2}\right) \tag{4}$$
for which there is a recursion relation etc. In maple you can convert hypergeometric functions to Jacobi polynomials and  vice versa and I expect mathematica does something similar. I appreciate this does not instantly give you the asymptotics but I was thinking you might consider it a  "reduction" (wikipedia cites a "Darboux formula" for the asymptotics of Jacobi polynomials at large order).
You also have the following relation, if:
$$I_{{n}} \left( \alpha \right) =\int _{-1/2\,\pi }^{1/2\,\pi }\!
 \left( \alpha+\sin \left( x \right)  \right) ^{n} \left( \cos \left( 
x \right)  \right) ^{2}{dx}$$
then:
$${\frac {d}{d\alpha}}I_{{n}} \left( \alpha \right) =nI_{{n-1}} \left( 
\alpha \right)$$ 
A: Look for the maximum value of $f(x)=(\alpha+\sin x)^n\cos^2x$.  For large values of $n$, it will be near $\pm\pi/2$.
Let $x=\pi/2-y$.  $y$ is going to be small, and then use Taylor series.
Take logs of $f(x)$, and differentiate, I get $$\frac{n\cos x}{\alpha+\sin x}-\frac{2\sin x}{\cos x}=0$$
Rearrange, to get $$n\sin^2y=2\cos y(\alpha+\cos y)$$
Expand in Taylor series: the leading term on the LHS is $ny^2$, on the RHS is $2(\alpha+1)$.
So I expect the maximum to be near $y=n^{-1/2}\sqrt{2\alpha+2}$.
Calculate the second derivative of $\log f(x)$ to find the width of the spike.
I get $4\sqrt{\pi}(\alpha+1)^{n+3/2}n^{-3/2}/e$.
A: I was too lazzy ! Use the binomal expansion and replace Cos[x]^2 by (1 - Sin[x]^2). You then have two almost identical integrals, the general term in the series being Sin[x]^m. The result of its integration between -Pi/2 and Pi/2 is given by  
(1+(-1)^m Sqrt[Pi] Gamma[(1+m)/2]) /(m Gamma[m/2])  
Is this of any interest ?  
Continuing the work, if a is real and positive, then the result of the integral is
(Pi/2) a^n Hypergeometric2F1[(1-n)/2,-n/2,2,1/a^2]
A: This is an interesting problem for sure. Concerning the integral values, the formulas look nice; they are Pi muliplied by a polynomial of alpha (of degree n). The coefficient of a^n is (1/2). The constant term in the polynomial is zero for odd values of n. In n is even, only even powers of a appear, if n is odd, only odd powers of n appear. I have not been able to establish recurrence relations for the coefficients.  
I know this is not the full answer to your question, but may be this could help you finding some tracks. In any manner, thanks for the problem !
