I want to know whether it is possible to show that $\displaystyle\int^{\frac{\pi}{2}}_{0}\cos{t}\cdot U_{2n-1}(\cos{t})\,dt=\frac{\pi}{2}$ for $n\in\mathbb{N}$,where $U_n$ is the Chebyshev polynomial of second kind.
I define $X(y,z)=\displaystyle\int^{\frac{\pi}{2}}_{0}\cos^{y}{t}\cdot U_{z}(\cos{t})\,dt$.
$X(m,2n-m)=2X(m+1,2n-m-1)-X(m,2n-m-2)$.
$$\begin{array}{c} X(1,1)=\frac{\pi}{2} & X(1,3)=\frac{\pi}{2} & \cdots \\ X(2,0)=\frac{\pi}{4} & X(2,2)=\frac{\pi}{2} & X(2,4)=\frac{\pi}{2} & \cdots \\ X(3,1)=\frac{3\pi}{8} & X(3,3)=\frac{\pi}{2} & X(3,5)=\frac{\pi}{2} & \cdots \\ X(4,0)=\frac{3\pi}{16} & X(4,2)=\frac{7\pi}{16} & X(4,4)=\frac{\pi}{2} & X(4,6)=\frac{\pi}{2} & \cdots \\ X(5,1)=\frac{5\pi}{16} & X(5,3)=\frac{15\pi}{32} & X(5,5)=\frac{\pi}{2} & X(5,7)=\frac{\pi}{2} & \cdots \\ X(6,0)=\frac{5\pi}{32} & X(6,2)=\frac{25\pi}{64} & X(6,4)=\frac{31\pi}{64} & X(6,6)=\frac{\pi}{2} & X(6,8)=\frac{\pi}{2} & \cdots \\ \vdots \\ \end{array}$$
It seems that $X(y,y-2)=\frac{(2^{y-1}-1)\pi}{2^{y}}$ for $y \ge 2$. If this is true then $X(y,y)=2X(y+1,y-1)-X(y,y-2)=\frac{\pi}{2}$, and $X(y,y+2k)=\frac{\pi}{2}$, so $X(1,2n-1)=\frac{\pi}{2}$ for any $n \ge 1$.
I tried to use $X(2k,0)=\displaystyle\int^{\frac{\pi}{2}}_{0}\cos^{2k}{t}\,dt=\frac{2\pi}{4^{k+1}}\binom{2k}{k}$, as well as $X(2k+1,1)=2X(2k+2,0)$. However they don't seem to simplify nicely, so I'm stuck here.
Original problem was to prove $\displaystyle\int^{\pi}_{0}\sin{nt}\cot{\frac{t}{2}}\,dt$ is an arithmetic progression. I know that with basic trigonometric identities it's obvious that their common difference is $0$, so what I want to prove is definitely true, but I'm interested in a method using the properties of the Chebyshev polynomials instead.
Define $$ I_n:=\int_0^{\pi/2} \cos{y} \,\, U_{2n-1}(\cos{y}) \, dy $$ We'll prove $I_n = \pi/2$ by a generating function. Take the odd bisection of the generating function $$ \sum_{n=0}^\infty U_n(x) t^n = \frac{1}{1-2 \, t\,x+t^2} $$ and we find $$ \sum_{n=1}^\infty U_{2n-1}(x) t^{2n-1} = \frac{1}{2} \Big(\frac{1}{1-2 \, t\,x+t^2} - \frac{1}{1+2 \, t\,x+t^2} \Big).$$ Therefore, $$\sum_{n=1}^\infty t^{2n} I_n = \frac{t}{2} \int_0^{\pi/2} dy \cos{y} \Big(\frac{1}{1-2 \, t\,\cos{y}+t^2} -\frac{1}{1+2 \, t\,\cos{y}+t^2} \Big) $$ $$= \frac{-1}{4} \int_0^{\pi/2} dy \cos{y} \Big(\frac{1}{\cos{y} - r}+ \frac{1}{\cos{y} + r}\Big) \, ,\quad r=\frac{1}{2}(t+\frac{1}{t})$$ $$ =\frac{1}{2} \int_0^{\pi/2} dy \, \frac{\cos^2{y}}{r^2-\cos^2{y}} = \frac{1}{2} \frac{\pi}{2} \big(\frac{r}{\sqrt{r^2-1}}-1\big) $$ where in the last step the integral has been performed in Mathematica. Since $t$ can be assumed to be $0<t<1,$ then $r>1.$ (This is mentioned because the form Mathematica spits out needs some simplification, and the branch needs to be correct.) Substitute the relation between $r$ and $t,$ simplify and we get $$ \sum_{n=1}^\infty t^{2n} I_n = \frac{\pi}{2}\frac{t^2}{1-t^2} = \frac{\pi}{2} \sum_{n=1}^\infty t^{2n} .$$ Equating coefficients of $t^2$ proves the OP's conjectured relation.