Integrals involving irrational numbers 
$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$

How should I approach getting an estimate on the numeric value in this case?
 A: Due to the symmetry of the graph of $\sin^{100} x$, the integral equals $$\frac{4}{2\pi} \int_0^{\pi/2} \sin^{100} x \ dx =\frac{2}{\pi} \cdot \frac{100!}{(2^{50}50!)^2}\cdot \frac{\pi}{2}= \frac{100!}{2^{100} \cdot (50!)^2}$$
using this result.
A: Probably, you should use complex numbers.
Say,
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \sin^{100}(x) dx =
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{(e^{ix} - e^{-ix})^{100}}{2^{100}} dx =
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}}(e^{2 ix} - 1)^{100} dx.
$$
If we exand $(e^{2 ix} - 1)^{100}$ and start doing integrals, we will see that they all are zero except
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}} {100\choose 50} e^{2 \times 50 \times ix} dx =
\frac{1}{2^{100}} {100\choose 50} \approx 0.0795...
$$
A: Integrate by parts successively to reduce the integral
$$I_{100}= \int_{0}^{2\pi} \sin^{100} x dx
=-\frac1{100}\int_{0}^{2\pi} \tan^{99} x \>d(\cos^{100}x)\\=\frac{99}{100}I_{98}  =\frac{99}{100} \frac{97}{98} I_{96}= \cdots =\frac{99}{100} \frac{97}{98} \cdots \frac12\cdot2\pi
$$
A: Use Mathematica [1]:
$$ \frac{1}{2\pi} \int_0^{2 \pi}\sin^{100}x\,  dx = \frac{12611418068195524166851562157}{158456325028528675187087900672}\approx 0.0795892$$
[1] https://www.wolframalpha.com/input/?i=Integrate%5BSin%5E100+%28x%29%2C%7Bx%2C0%2C2Pi%7D%5D%2F%282Pi%29
