if$$\alpha_m = \int^{1}_{-1}\cos \pi x (x^2-1)^m dx$$
then how do I get
$$\alpha_m = \frac{-1}{\pi^2}(2m(2m-1)\alpha_{m-1} + 4m(m-1)\alpha_{m-2})$$
I have to integrate by parts!
But $$\alpha_n = -\frac{1}{\pi^2}\int^{1}_{-1}\cos \pi x (m(m-1)(x^2-1)^{m-2}4x^2 + 2m(x^2-1)^{m-1})dx$$
I can see how I might be getting closer, but these factors of $x$ are not doing me any favours.
Integrating by parts once we get: $$ \alpha_m = \int_{-1}^{1}\frac{\sin(\pi x)}{\pi}\frac{d}{dx}(x^2-1)^m\,dx = \frac{m}{\pi}\int_{-1}^{1}2x\sin(\pi x)\cdot(x^2-1)^{m-1}\,dx$$ and by integrating by parts twice, through $\int 2x\sin(\pi x)\,dx = \frac{2\sin(\pi x)}{\pi^2}-\frac{2x\cos(\pi x)}{\pi}$, we get: $$ \alpha_m = \frac{m(m-1)}{\pi}\left(\frac{2}{\pi^2}\int_{-1}^{1}2x\sin(\pi x)(x^2-1)^{m-2}\,dx-\frac{4}{\pi}\int_{-1}^{1}x^2\cos(\pi x)(x^2-1)^{m-2}\,dx\right)$$ where the first term in the RHS depends on $\alpha_{m-1}$ and the second term, by writing $x^2$ as $(x^2-1)+1$, depends on $\alpha_{m-1}$ and $\alpha_{m-2}$.
