# How to prove that $\int_0^{2\pi} \sqrt{1+\cos^2{t}}\; dt>2\pi$?

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

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

## 2 Answers

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

• 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
• @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 $(2 \pi, 0)$.

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