1. ## An inequality

The upper bound given in this problem is a good approximation for the integral:

Problem: Prove that $\displaystyle \int_0^1 e^{\cos x} \ dx \leq e \sqrt{2} \tan^{-1}(\sqrt{2}/2).$

2. By the Rectangle Rule, $\displaystyle \int_{0}^{1} e^{\cos x} \ dx \approx e^{\cos \frac{1}{2}}$. Or we know that $\displaystyle e^{\cos x}$ has a bounded derivative over $\displaystyle [0,1]$. Now by the MVT, $\displaystyle xf'(y_x) = f(x)-f(0)$ for some $\displaystyle y_x \in [a,x]$. Also, $\displaystyle x<1$. Then

$\displaystyle \left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| = \left|\int_{0}^{1} x f'(y_x) \ dx \right|$

$\displaystyle \left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| \leq \frac{1}{2} \sup_{0 \leq x \leq 1} |-e^{\cos x} \sin x|$

3. OK, I have come up with an incredibly contrived solution suggested by the answer. Clearly it is equivalent to

$\displaystyle \int_0^1\mathrm e^{\cos x-1}\mathrm dx\leq\int_0^1\frac 2{2+x^2}\mathrm dx$, so this suggests the inequality $\displaystyle \mathrm e^{\cos x-1}\leq\frac 2{2+x^2}$. If there's a better way...

Here we go. Start with $\displaystyle \sin x\leq x$ and $\displaystyle \sin^2 x+\cos x=1+\cos x(1-\cos x)\geq 1$ on $\displaystyle [0,1]$. Therefore $\displaystyle \sin x\leq x(\sin^2 x+\cos x)$.

This rearranges to $\displaystyle \frac{\sin x-x\cos x}{\sin^2 x}\leq x$. The LHS is the derivative of $\displaystyle \frac x{\sin x}$, so integrate both sides from $\displaystyle 0$ to $\displaystyle t$, where $\displaystyle t\in[0,1]$.

Thus $\displaystyle \frac t{\sin t}-1\leq \frac{t^2}2$ and so $\displaystyle \sin t\geq\frac t{1+t^2/2}$. Integrate this from $\displaystyle 0$ to $\displaystyle x$ and $\displaystyle 1-\cos x\geq \ln(1+x^2/2)$ pops up.

Guess what? $\displaystyle \cos x-1\leq\ln\left(\frac2{2+x^2}\right)$, and we arrive at our destination. Where's my hat? Rabbit, anyone?