I would like to know the proof of $$ \int_a^b \frac{T_n(x/a)T_n(x/b)\, dx}{x(b^2-x^2)^{1/2}(x^2-a^2)^{1/2}}=\frac{\pi}{2 ab}, 0<a<b, n \in \Bbb N $$ where $T_n(x)$ is the Chebyshev polynomial of order $n$.
Perhaps it may be possible to prove it by using the following recurrence relations. $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$
$(1-x^2){T_n}'(x)=nT_{n-1}(x)-nxT_n(x)$
$(1-x^2){T_n}''(x)-x{T_n}'(x)+n^2 T_n(x)=0$
Could anyone help me solve this? Thanks.