I read this quote attributed to VI Arnold. 


  "Who can't calculate the average value of the one hundredth power of the sine function within five minutes, doesn't understand mathematics - even if he studied supermanifolds, non-standard calculus or embedding theorems."


EDIT Source is "A mathematical trivium" A book of 100 problems that university students "should be able to solve". The statement asks for calculation within 10% accuracy. 



So the average value over the entire domain should be the same as the average value over  $[0,\pi/2]$

$$\langle\sin^{100} (x)\rangle= \frac{\int_0^{\pi/2} \sin^{100}(x) dx}{\int_0^{\pi/2} dx}.$$

So here's what I did:

First, this graph would be a train of highly sharp peaks. The integrand would assume values close to zero a up till before it sharply rises to 1. 

So up till some $\epsilon \in [0,\pi/2]$ we will have $\sin x \approx x$ and for the remaining $\pi/2 - \epsilon$ interval I could find the area of triangle with base $\pi/2 - \epsilon$ and height $1$

$$\langle \sin^{100} (x)\rangle \approx \frac{2}{\pi} \left(\int_0^\epsilon x^{100} dx + .5 (\frac{\pi}{2}-\epsilon)\right).$$

I believe in principal it should be possible to find an $\epsilon$ such that the above expression yields the exact answer. So I try to approximate it, no good. Then I try mathematica and it is looking like there is no $\epsilon$ for which the value I am expecting is even close to the actual value. I plot the original and find that my approximation is hopeless.

Not to mention that my 5 minutes were over. So I admit I do not understand mathematics and humbly ask if someone could:


Point out my mistake (Other than that $\epsilon$ is probably incomputable within 5 mins)
How the hell is this done in 5 minutes?


The picture below has the $\sin^{100} x$ in blue (bottom) and my approximation of it plotted against $\epsilon$ (pink). Although there is no reason for them to be together, the upper graph has a minima quite above the exact value of the integral. 

EDIT
Just realized

Let $$u=\cos x.$$

$$\int_0^{\pi/2} \sin^{100}(x) dx = \int_0^1 (1-u^2)^{99/2}du\approx \int_0^1 \left(1 - \frac{99}{2} u^2\right) du $$

Since $$\sin x=\frac{e^{ix}-e^{-ix}}{2i} $$
and, for $k\in\mathbb Z $, $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&k\ne0\\ 2\pi&k=0\end{array}\right.,$$
we have $$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi,$$ and the average value is $$\frac{\binom{100}{50}}{2^{100}}.$$