Find the average of $\sin^{100} (x)$ in 5 minutes? 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 $, $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&amp;k\ne0\\ 2\pi&amp;k=0\end{array}\right.,$$
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 $$\frac{\binom{100}{50}}{2^{100}}.$$
I tended to the five minutes side, less the accuracy.  As shown by others, if you are on top of your complex variables the exact answer (unevaluated) can be found in under five minutes, and computers can handle the problem in a whiz.  

It was fun to consider the intuitive shape of the equation.  The average over a large region is simply $1/2$; near zero, it is essentially zero.  However, an accuracy to a percent takes a little bit of work.

Approximate the integral as an area of height 1 and width from where $ \sin^{100} (x)=.5$ to $ \pi /2 $.  In the region of interest, 
$$\sin(x)=1-(\pi/2-x)^2/2 $$ Sure, the sloping curve does not quite have rotational symmetry, but at 10% a rectangle should be fine.  

If $\sin^{100}(z)=.5$, then $\sin(z)=\sqrt[100]{.5} $, which I am not used to finding without a calculator.  I took $\sqrt{.49}=.7$ and $\sqrt{.7} \approx 1-.3/2 -.01 = .84$ because I was not quite near enough to one to assume the number was constant, that is, $(1-a)^n \approx 1-n\cdot a$, 

Similarly, $\sqrt[25]{.84} \approx 1-\dfrac{\frac{16}{100}}{25}$, and for the width of my rectangle, y, $$\frac{1}{2} y^2 =  \dfrac{\frac{16}{100}}{25}$$ $$(10 y)^2 = \frac{32}{25}$$ $$10y\approx 1+ \frac{1}{2} \frac{7}{25}$$ $$10y\approx 1+.04\cdot 3.5$$ or $y\approx .114$, giving an average of $1\cdot y\frac{2}{\pi}\approx .075$

I assume the primary error comes from the approximation finding the 25th root, which should be less than the estimate, so $y$ should be larger than found.  I partially accommodated this in my lazy division by $\pi$ in my head.