Integral involving Cosine functions The following integral arises in a problem in elasticity:
Let $n$ be a positive integer.
Evaluate
\begin{equation*}
I=\int_0^{\pi}\frac{\cos(3\eta/4)\cos n\eta\,d\eta}{(1+\cos\eta)^{1/4}}.
\end{equation*}
Mathematica is able to evaluate this integral for $n=1$, $2$, $3$ etc., but does not evaluate it for arbitrary $n$ (as a function of $n$).
For example,
\begin{align*}
I&=\frac{35\pi 2^{-3/4}}{128}, && (n=5), \\
&=-\frac{63\pi 2^{-3/4}}{256}, && (n=6).
\end{align*}
 A: Here's a proof that 
$$I_n:=\int_0^\pi \frac{\cos(3x/4)\,\cos(nx)}{(1+\cos{x})^{1/4}} dx = 2^{-3/4} \pi \, n \binom{1/2}{n}\quad , \quad n=0,1,2,...$$
where the $()$ is a binomial symbol.
By a half-angle formula and an integral scale we find
$$I_n=2^{-1/4} \int_0^{\pi/2} \frac{\cos(3x/2)\,\text{T}_{2n}(\cos{x})}{(\cos{x})^{1/2}} dx $$ where the Chebyshev polynomial of the first kind has been used (T$_n$(cos x)=cos(n x) ),
With another half-angle trig ID use,
$$I_n=2^{-3/4} \int_0^{\pi/2}\frac{\cos{x}\sqrt{1+\cos{x}}-\sin{x}\sqrt{1-\cos{x}}}
{\sin{x} \sqrt{\cos{x}}}
\text{T}_{2n}(\cos{x}) \sin{x}dx 
$$
$$=2^{-3/4}\int_0^1\Big(\frac{\sqrt{y}}{\sqrt{1-y}} - \frac{\sqrt{1-y}}{\sqrt{y}}\Big)
\text{T}_{2n}(y)\, dy$$
where the change of variables $y=\cos{x}$ has been made.
Integrate by parts once to get 
$$I_n=2^{-3/4}\, 4n\,\int_0^1\sqrt{y(1-y)}\,\text{U}_{2n-1}(y) \, dy $$
where the Chebyshev polynomial of the second kind has been used.  We find this last integral in closed form for $n=1,2,...$ by using the generating function
$$ \sum_{k=0}^\infty U_k(y)t^k = \frac{1}{1-2\,t\,y + t^2} .$$
(Actually, we will take the odd bisection.)
$$J_n:=\int_0^1\sqrt{y(1-y)}\,\text{U}_{2n-1}(y) \, dy =[t^{2n-1}]\sum_{k=1}^\infty t^{2k-1} \int_0^1 U_{2k-1}(y) \sqrt{y(1-y)} \,dy$$
where the square brackets are the 'coefficient of' operator. With the generating function inserted,
$$J_n:=\frac{1}{2}[t^{2n-1}] \int_0^1 \Big(\frac{1}{1-2\,t\,y + t^2} - \frac{1}{1+2\,t\,y + t^2} \big)\sqrt{y(1-y)} \,dy$$
It can be proved (Mathematica) that
$$ \int_{0}^1 \frac{\sqrt{y(1-y)}}{1-2\,t\,y+t^2} dy= \frac{\pi}{4t^2}\big(1+(t-1)(t+\sqrt{1+t^2})\big) $$
so it follows that
$$J_n:=\frac{\pi}{2}[t^{2n-1}]\frac{2t(\sqrt{1+t^2}-1)}{4t^2}=
\frac{\pi}{4}[t^{2n-1}]\sum_{n=1}^\infty \binom{1/2}{n}t^{2n-1}$$
$$=\frac{\pi}{4} \binom{1/2}{n}.$$
Algebra completes the proof for $n=1,2,3...$  For $n=0,$ do the integral explicitly,
and it will shown to be consistent with the given formula.
