Prove that $\int_1^a \frac{T_n(x) T_n(x/a)}{\sqrt{a^2 - x^2} \sqrt{x^2 - 1^2}} \frac{a}{x} \mathrm{d}x = \frac{\pi}{2}$ In the paper, Representation of a Function by Its Line Integrals, with Some Radiological Applications, A. M. Cormack, Journal of Applied Physics 34, 2722 (1963), an integral identity is expressed which can be reduced to:
$$I_n(a) = \int_1^a \frac{T_n(x)  T_n(x/a)}{\sqrt{a^2 - x^2} \sqrt{x^2 - 1^2}}  \frac{a}{x} \mathrm{d}x = \frac{\pi}{2}$$
where $T_n(x)$ is the $n^\text{th}$ order Chebyshev polynomial of the first kind.
I'm not quite sure where this result comes from, and was looking to see how it could be derived. The paper notes that it can be shown that
$$I_{n+1} = I_{n-1}, \quad I_0 = I_1 = \frac{\pi}{2}$$
from which then it is apparent that $I_n(a) = \frac{\pi}{2}$
However, I'm not sure where the $I_{n+1} = I_{n-1}$ comes from.
Attempting to substitute the recurrence relationship
$$T_n(x) = \frac{T_{n-1}(x) + T_{n+1}(x)}{2x}$$
seemed like a good start, but ends up converting the integral to a form that's different from $I_{n-1}(a)$ and $I_{n+1}(a)$, and also generates some unwanted cross terms.
Could anyone suggest a push in the right direction?
 A: Using the trigonometric definition of the Chebyshev polynomial of the first kind we can write:
$$\small  I_{n}(a)=\int_1^a \frac{T_n(x) T_n\left(\frac{x}{a}\right)}{\sqrt{x^2 - 1}\sqrt{a^2 - x^2}}\frac{a}{x}dx=\int_1^a \frac{\cosh(n\operatorname{arccosh} x) \cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{x^2 - 1}\sqrt{1 - \frac{x^2}{a^2}}} \frac{dx}{x}$$
$$I_0(a)=\int_1^a \frac{1}{\sqrt{x^2 - 1}\sqrt{1 - \frac{x^2}{a^2}}} \frac{dx}{x}\overset{x\to \frac{a}{x}}=\int_1^a \frac{\frac{x}{a}\cdot x}{\sqrt{1-\frac{x^2}{a^2}}\sqrt{x^2-1}}\frac{dx}{x}=I_1(a)$$
$$\overset{1-\frac{x^2}{a^2}=t^2}=\int_0^{\sqrt{1-\frac{1}{a^2}}}\frac{1}{\sqrt{1-\frac{1}{a^2}-t^2}}dt=\arcsin\left(\frac{t}{\sqrt{1-\frac{1}{a^2}}}\right)\bigg|_0^{\sqrt{1-\frac{1}{a^2}}}=\frac{\pi}{2}$$
So as the paper mentions, we only need to show that $I_{n-1}(a)=I_{n+1}(a)$ to obtain $I_{n}(a)=\frac{\pi}{2}$.

$$\small T_{n-1}\left(x\right)=\cosh((n-1)\operatorname{arccosh} x)=\color{red}{x\cosh(n\operatorname{arccosh} x)}-\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)}$$
$$\small T_{n-1}\left(\frac{x}{a}\right)=\cos\left((n-1)\arccos \left(\frac{x}{a}\right)\right)=\color{blue}{\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}+\color{red}{\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}$$
$$\small T_{n+1}\left(x\right)=\cosh((n+1)\operatorname{arccosh} x)=\color{red}{x\cosh(n\operatorname{arccosh} x)}+\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)}$$
$$\small T_{n+1}\left(\frac{x}{a}\right)=\cos\left((n+1)\arccos \left(\frac{x}{a}\right)\right)=\color{blue}{\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}-\color{red}{\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}$$
When we rewrite the numerator of $I_{n-1}(a)-I_{n+1}(a)$ (we are trying to show that this vanishes) using the above identities, only the terms of the same color remains (twice), giving:
$$\frac12\left(T_{n-1}(x)T_{n-1}\left(\frac{x}{a}\right)-T_{n+1}(x)T_{n+1}\left(\frac{x}{a}\right)\right)$$
$$\small = \color{red}{x\cosh(n\operatorname{arccosh} x)\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}-\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}$$

So $ I_{n-1}(a)-I_{n+1}(a)$ takes the form of:
$$\small 2\color{chocolate}{\int_1^a \frac{\cosh(n\operatorname{arccosh} x)\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{x^2-1}}dx}-\frac2a\color{green}{\int_1^a\frac{\sinh(n\operatorname{arccosh} x)\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{1-\frac{x^2}{a^2}}}dx}$$
$$\small \color{chocolate}{\int_1^a \left(\frac{\sinh(n\operatorname{arccosh} x)}{n}\right)'\sin\left(n\arccos \left(\frac{x}{a}\right)\right)dx}\overset{IBP}=\frac1a\color{green}{\int_1^a\frac{\sinh(n\operatorname{arccosh} x)\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{1-\frac{x^2}{a^2}}}dx}$$
$$\Rightarrow I_{n-1}(a)-I_{n+1}(a)=0\Rightarrow I_{n-1}(a)=I_{n+1}(a)$$
