User:Dan Nessett/Sandboxes/Sandbox 2

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

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

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

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

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

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

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