Use the residue theorem to compute the real integral: $$I = \int_{0}^{2\pi} \sin^{2n}\theta d\theta$$
I have considered a contour around a unit circle C, and used the substitutions: $sin\theta = \frac{1}{2i} (z-\frac{1}{z})$ and $d\theta = -\frac{i}{z} dz$ to get:
$$I = \frac{-i}{(2i)^{2n}} \oint_C \frac{(z^2 -1)}{z}^{2n} \frac{dz}{z}$$
$$I = \frac{-i}{(2i)^{2n}} \oint_C \frac{{(z^2 -1)}^{2n}}{z^{2n+1}} dz$$
I am now a bit confused about how to calculate the residue at the pole of order $2n+1$ at $z=0$
I believe I must compute: $$Res(z=0) = \frac{-i}{(2n)!(2i)^{2n}} \lim_{z \to 0} \frac{d^{2n}}{dz^{2n}}(z^2-1)^{2n} $$
But I'm not sure how to go about this? Any help would be greatly appreciated!
