Use ∫02πsinnxdx=nn−1∫02πsinn−2xdx
to show that, for odd powers of sine,
∫02πsin2n+1xdx=3⋅5⋅7⋅⋅⋅⋅⋅(2n+1)2⋅4⋅6⋅⋅⋅⋅⋅2n
Attempt:
I subbed in 2n + 1 into n in the original reduction formula and got:
∫02πsin2n+1xdx=2n+12n∫02πsin2n−1xdx
but I cant see what more I can do
to show that, for odd powers of sine,
∫02πsin2n+1xdx=3⋅5⋅7⋅⋅⋅⋅⋅(2n+1)2⋅4⋅6⋅⋅⋅⋅⋅2n
Attempt:
I subbed in 2n + 1 into n in the original reduction formula and got:
∫02πsin2n+1xdx=2n+12n∫02πsin2n−1xdx
but I cant see what more I can do