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Integral evaluation

Posted November 19th, 2007 by Anonymous
in

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

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Answer

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

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

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