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Let $$I=\int_0^{2\pi} \sqrt{1+\cos^2{(t)}}\;dt$$

How to prove, in an elementary way, that $I>2\pi$?

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2 Answers 2

up vote 10 down vote accepted

Here is a hint: $$ 1+\cos^2(t)\geq 1 $$ for all $t$.

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And another: there exists $t\in [0,2\pi]$ such that $1+\cos^2(t) > 1$ and $1+\cos^2(t)$ is continuous. –  AlexR Aug 15 '13 at 17:47
    
@AlexR Yours is the hint. –  Git Gud Aug 15 '13 at 17:48
    
@GitGud combined with the one above and elementary properties of the Integral, you have the complete proof for the strict inequality –  AlexR Aug 15 '13 at 17:49
    
Or you can use geometrical approach. Integral is area under graph ... –  Antoine Aug 15 '13 at 17:59
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@Git Gud Strictly above a line $y=1$ I think. But it was just a hint anyway ... –  Antoine Aug 15 '13 at 18:03

For a geometric approach, note that the integral $I$ is the arc length of the function $y = \sin x$ on the interval $[0, 2\pi]$. The inequality follows since drawing a straight line from $(0,0)$ to $(2 \pi, 0)$ is strictly shorter than the route the graph of $\sin x$ takes from $(0,0)$ to $(2 \pi, 0)$.

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Excellent. Then there are lower bounds by drawing polygons through known points on the curve. –  zyx Aug 15 '13 at 18:40

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