TITLE: Where is the mistake in my integration of $\sin^{2n}(\theta)$?
QUESTION [6 upvotes]: Note: it's most likely in steps $3-6$.
$1.$ Deriving a useful formula:
$$-i e^{i2\theta}+i=-i \cos(2 \theta)+\sin (2 \theta)+i=2 \sin(\theta) \cos (\theta)+i(1-\cos (2 \theta))$$
$$=2 \sin (\theta)\cos(\theta)+i  2\sin^2 (\theta)=2 \sin (\theta)(\cos(\theta)+i \sin (\theta))=2 \sin (\theta) e^{i \theta}$$
$2.$ Now, exponentiating to $2n$:
$$2^{2n} \sin^{2n} (\theta) e^{i 2n \theta}=i^{2n}(- e^{i2\theta}+1)^{2n}$$
$$=\sum_{k=0}^{n}\binom{\color{red}{2n-k}}{k} (i)^{2n} (-1)^k  e^{ik2\theta}$$
$3$. Dividing by $e^{i2n \theta}$ (as cannot $=0$):
$$2^{2n} \sin^{2n}(\theta)=\sum_{k=0}^{n}\binom{2n-k}{k}(-1)^k(i)^{2n} e^{i(k-n)2\theta}$$
$4.$Now, to the integral:
$$\int_{0}^{\pi}\sin^{2n}(\theta) d \theta=\frac{(i)^{2n}}{2^{2n}} \int_{0}^{\pi} \sum_{k=0}^{n}  \binom{2n-k}{k} (-1)^k e^{i(k-n)2\theta}d \theta$$
$5.$Note that $$\int_{0}^{\pi}e^{iN2 \theta}d \theta=0 \text{ (if N }(\in \mathbb{Z}) \ne 0)$$
$6.$So only $k=n$ contributes nonzero value to the integral:
$$\int_{0}^{\pi}\sin^{2n}(\theta) d \theta=\frac{(-1)^n(i)^{2n}}{2^{2n}} \int_{0}^{\pi} \binom{2n-n}{n} e^{i(n-n)2\theta}d \theta$$
$$=\frac{(-1)^n(-1)^n}{2^{2n}} \binom{n}{n} \pi=\frac{1}{2^{2n}} \pi$$
Where have I gone wrong?

REPLY [4 votes]: $$
\begin{align}
\sin^{2n}(\theta)
&=\left(\frac1{2i}\right)^{2n}\left(e^{i\theta}-e^{-i\theta}\right)^{2n}\\
&=\left(-\frac14\right)^n\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-k)\theta}e^{-ik\theta}(-1)^k\\
&=\left(\frac14\right)^n\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-2k)\theta}(-1)^{n-k}
\end{align}
$$
As you say, the only term that contributes is $n=k$, so the integral is
$$
\frac\pi{4^n}\binom{2n}{n}
$$