MHB Prove Inequality: Integral of Square Root vs. Trigonometric Function

[anemone]

Here is this week's POTW:

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Show that for every $0<\theta \le \pi$, one has $\displaystyle \int_0^\theta \sqrt{1+\cos^2 t} \,dt>\sqrt{\theta^2+\sin^2 \theta}$.

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[anemone]

Congratulations to lfdahl for his correct solution (Cool) , which you can find below:

Spoiler

With the simple parameterization: $x(t) = t$, $y(t) = \sin t$ -

the arc length on the sine curve from Origo to the point $(\theta, \sin \theta)$, where $0 < \theta \leq \pi$ can be expressed exactly as the integral given in the problem: \[\int_{0}^{\theta }\sqrt{1+\cos^2t}\, dt\]

Now consider the triangle with the corners $(0,0)$, $(\theta,0)$ and $(\theta, \sin \theta)$. The hypotenuse of the triangle is a chord of length $\sqrt{\theta^2 + \sin^2\theta}$ on the graph of the sine function. Since sine is concave on the interval given, we know that every chord is shorter, than the arc length of sine. Thus, we can conclude, that

\[\int_{0}^{\theta }\sqrt{1+\cos^2t}\, dt > \sqrt{\theta ^2+\sin^2\theta }\]