Hi
it would be really great if someone could help me with the following problem. I seem to have a great thick fog descending and cannot find my way out!!
Prove that:
(pi^3)/48 <= integral [0,pi/2] x^2/(2-sinx) dx <= (pi^3)/24
On $\displaystyle \left[0,\frac{\pi}{2}\right] 0 \le \sin x \le 1 $ so
$\displaystyle
-1 \le - \sin x \le 0
$
$\displaystyle
1 \le 2 - \sin x \le 2
$
$\displaystyle
\frac{1}{2} \le \frac{1}{2- \sin x} \le 1
$
$\displaystyle
\frac{x^2}{2} \le \frac{x^2}{2- \sin x} \le x^2
$
Now integrate over the interval.