Integral evaluation

Posted November 19th, 2007 by Anonymous

Prove that
$\int_{0}^{1}\frac{\arctan x}{\sin x}dx>\frac{8}{9}$

Comments

On November 22nd, 2007 Isoscel says:

On $(0,1]$ we have $\arctan x\geq x-\frac{x^3}{3}$ and $\sin x\leq x$ so $\frac{1}{\sin x}\geq \frac{1}{x}$ Finally we have

$\frac{\arctan x}{\sin x}\geq 1-\frac{x^2}{3}$

and also

$\int_0^1 \frac{\arctan x}{\sin x}dx >\int_0^1(1-\frac{x^2}{3})dx=1-\frac{1}{9}=\frac{8}{9}$