# Finding $\int_{0}^{1} \frac{T_{i}^{*''}(x) T_{j}^{*}(x)}{\sqrt{x-x^2}} dx$

Shifted Chebyshev polynomials $$T_{i}^{*}(x) = \cos(i \arccos(2x-1))$$

We want to calculate $$I=\int_{0}^{1} \frac{T_{i}^{*''}(x) T_{j}^{*}(x)}{\sqrt{x-x^2}} dx$$ Which is equal to $$\sum_{\substack{i=0 \\ (i+j) even}}^{j-2} 2 \ \delta_{ij}\ j(j^2-i^2)c_i$$

How to get this result

We know that $$\int_{0}^{1} \frac{T_{i}^{*}(x) T_{j}^{*}(x)}{\sqrt{x-x^2}} dx = \delta_{ij}$$ Where $$\delta_{ij} = \left\{ \begin{array}{ll} \frac{\pi}{c_{i}} & i=j \\ 0 & i\neq j \end{array} \ \ \ , c_{i} = \left\{ \begin{array}{ll} 1 & i=0 \\ 2 & i\geq 1 \end{array} \ \ \ \right. \right.$$ And We can write $$T_{i}^{*''}(x) =\sum_{k=0}^{i-2} a_{k} \ T_{i}^{*}(x)$$ So how to get the required result above ? if we got $$a_{k}$$ we are supposed to get the required result How to get the required result above ? And Is there a general formula to get $$a_k$$ such that $$T_{i}^{*(n)}(x) = \sum_{k=0}^{i-2} a_{k} \ T_{i}^{*}(x)$$ Using Mathematica I have got :