# Find the average of sin100(x)\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 $$[0,π/2][0,\pi/2]$$

$$⟨sin100(x)⟩=∫π/20sin100(x)dx∫π/20dx.\langle\sin^{100} (x)\rangle= \frac{\int_0^{\pi/2} \sin^{100}(x) dx}{\int_0^{\pi/2} dx}.$$

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 $$ϵ∈[0,π/2]\epsilon \in [0,\pi/2]$$ we will have $$sinx≈x\sin x \approx x$$ and for the remaining $$π/2−ϵ\pi/2 - \epsilon$$ interval I could find the area of triangle with base $$π/2−ϵ\pi/2 - \epsilon$$ and height $$11$$

$$⟨sin100(x)⟩≈2π(∫ϵ0x100dx+.5(π2−ϵ)).\langle \sin^{100} (x)\rangle \approx \frac{2}{\pi} \left(\int_0^\epsilon x^{100} dx + .5 (\frac{\pi}{2}-\epsilon)\right).$$

I believe in principal it should be possible to find an $$ϵ\epsilon$$ 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 $$ϵ\epsilon$$ 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:

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

The picture below has the $$sin100x\sin^{100} x$$ in blue (bottom) and my approximation of it plotted against $$ϵ\epsilon$$ (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 $$u=cosx.u=\cos x.$$

$$∫π/20sin100(x)dx=∫10(1−u2)99/2du≈∫10(1−992u2)du\int_0^{\pi/2} \sin^{100}(x) dx = \int_0^1 (1-u^2)^{99/2}du\approx \int_0^1 \left(1 - \frac{99}{2} u^2\right) du$$

and, for $k\in\mathbb Z$,