x = cos θ ⟹ d x = − sin θ d θ and 1 − x 2 = ( sin θ ) 2 ⟹ x 2 − 1 = − ( sin θ ) 2 . {\displaystyle 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}.}
( x 2 − 1 ) l = ( − 1 ) l ( sin θ ) 2 l {\displaystyle \left(x^{2}-1\right)^{l}\ =\ (-1)^{l}\left(\sin \theta \right)^{2l}}
( x 2 − 1 ) l − 1 = ( − 1 ) l − 1 ( sin θ ) 2 l − 2 {\displaystyle \left(x^{2}-1\right)^{l-1}\ =\ (-1)^{l-1}\left(\sin \theta \right)^{2l-2}}
∫ − 1 1 ( x 2 − 1 ) l d x = ( − 1 ) l + 1 ∫ 0 π ( sin θ ) 2 l + 1 d θ {\displaystyle \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 1 ( x 2 − 1 ) l − 1 d x = ( − 1 ) l ∫ 0 π ( sin θ ) 2 l − 1 d θ {\displaystyle \int \limits _{-1}^{1}(x^{2}-1)^{l-1}dx\ =\ (-1)^{l}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l-1}d\theta }
∫ 0 π sin n θ d θ = ( n − 1 ) n ∫ 0 π sin n − 2 θ d θ {\displaystyle \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 }
∫ − 1 1 ( x 2 − 1 ) l d x = ( − 1 ) l + 1 ∫ 0 π ( sin θ ) 2 l + 1 d θ = ( − 1 ) l + 1 2 l 2 l + 1 ( − 1 ) l + 1 ∫ 0 π ( sin θ ) 2 l − 1 d θ = − 2 l 2 l + 1 ∫ − 1 1 ( x 2 − 1 ) l − 1 d x {\displaystyle \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}
