[SOLVED] Inequality involving area under a curve

[anemone]

Prove that for every $x\in (0,\,1)$ the following inequality holds:

$\displaystyle \int_0^1 \sqrt{1+(\cos y)^2} dy>\sqrt{x^2+(\sin x)^2}$

 

[anemone]

Spoiler: Solution of other

Clearly $\displaystyle \int_0^1 \sqrt{1+(\cos y)^2} dy \ge \int_0^x \sqrt{1+(\cos y)^2} dy$ for each fixed $x\in (0,\,1)$. Observe that $\displaystyle \int_0^x \sqrt{1+(\cos y)^2} dy$ is the arc length of the function $f(y)=\sin y$ on the interval $[0,\,x]$ which is clearly strictly greater than the length of the straight line between the points $(0,\,0)$ and $(x,\, \sin x)$ which in turn is equal to $\sqrt{x^2+(\sin x)^2}$.