User:Dan Nessett/Sandboxes/Sandbox 2

I think it is useful to get on record the following derivation. It doesn't belong in the proof itself, but putting it on this talk page gets it into the history of the article:

$x=\cos \theta \;\Longrightarrow \;dx=-\sin \theta d\theta \quad {\hbox{and}}\quad 1-x^{2}=(\sin \theta )^{2}\Longrightarrow \;x^{2}-1=\ -(\sin \theta )^{2}.$

$\left(x^{2}-1\right)^{l}\ =\ (-1)^{l}\left(\sin \theta \right)^{2l}$

$\left(x^{2}-1\right)^{l-1}\ =\ (-1)^{l-1}\left(\sin \theta \right)^{2l-2}$

$\int \limits _{-1}^{1}(x^{2}-1)^{l}dx\ =\ (-1)^{l+1}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta$

$\int \limits _{-1}^{1}(x^{2}-1)^{l-1}dx\ =\ (-1)^{l}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l-1}d\theta$

$\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta$

$\int \limits _{-1}^{1}(x^{2}-1)^{l}dx\ =\ (-1)^{l+1}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta \ =\ (-1)^{l+1}{\frac {2l}{2l+1}}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l-1}d\theta \ =\ -\ {\frac {2l}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$