Show that ${3 \over 2} \log{5} \leq \int_0^2{{x \over \sin{x}}d{x}} \leq \sqrt{6}\log{(\sqrt{2}+\sqrt{3})}$

Question

Show that ${3 \over 2} \log{5} \leq \int_0^2{{x \over \sin{x}}d{x}} \leq \sqrt{6}\log{(\sqrt{2}+\sqrt{3})}.$

I can show that ${1 \over 2} \log{5} \leq \int_0^2{{x \over \sin{x}}d{x}}$ but not this. And for the right side, I have no idea. Can anyone give any hints?

Answer 1

Observe on $[0, 2]$ we see that \begin{align} x-\frac{x^3}{3!} \leq \sin x \leq x-\frac{x^3}{9} \end{align} which means \begin{align} \frac{9}{9-x^2}\leq \frac{x}{\sin x} \leq \frac{6}{6-x^2}. \end{align} Then we see that \begin{align} \frac{3}{2}\log 5=9\int^2_0 \frac{1}{9-x^2}\ dx\leq \int^2_0 \frac{x}{\sin x}\ dx \leq 6\int^2_0 \frac{dx}{6-x^2} = \sqrt{6}\log(\sqrt{2}+\sqrt{3}). \end{align}