Explanation of $\int_0^{2\pi}\sin^{100}x\,dx$. I was browsing this thread when I came across this answer. I can neither make heads nor tail of it. Can someone help me understand it?
This I understand:
$$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&k\ne0\\ 2\pi&k=0\end{array}\right.,$$
But how does he get this? Where does the summation come from?
$$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi.$$
EDIT: The answer is so obvious...I can't believe I didn't notice it! I kept expanding $\sin^{100}(x)$ as $\frac{1}{2i} (e^{100ix} + e^{-100ix})$ which got me nowhere.
 A: We have
$$\sin x=\frac1{2i}(e^{ix}-e^{-ix})$$
so by the binomial formula we get
$$\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{100-k}\binom{100}ke^{kix}e^{-ix(100-k)}$$
so
$$\int_0^{2\pi}\sin^{100}xdx=\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{k}\binom{100}k\underbrace{\int_0^{2\pi}e^{(2k-100)ix}dx}_{=2\pi\;\text{only for}\; k=50}=\frac{2\pi}{2^{100}}\binom{100}{50}$$
A: Hint: Using the binomial theorem
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}\implies \sin^{100}x=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}ke^{kix}e^{-ix(100-k)}\;\ldots$$
A: Write 
$$
\sin x = \frac{e^{ix} - e^{-ix}}{2i}
$$
Then 
$$
\sin^{100} x  = \frac{(a + b)^{100}}{2^{100}}
$$
where $a = e^{ix}$ and $b = -e^{-ix}$.
Now use the binomial theorem:
$$
(a + b)^{100} = \sum_{k=0}^{100}\left( \begin{array}{c} 100 \\ k \end{array}  \right) a^k b^{100-k}
$$
and the first summation step in the formula you were puzzled about you give falls out.
And then in that integral,  only the term where the factor in front of $x$ in the exponent gives a non-zero integral.  So only the term with $k - (100-k) = 0$, or $k=50$, is non-zero. That is where the $\left( \begin{array}{c} 100 \\ 50 \end{array}  \right)$ comes from.
