How to prove that $\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int_0^{\pi/2} \frac{\sin x}{x(x+1)} \, \mathrm{d}x < \ln\frac{\pi+2}{2}$ How to prove that $$\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{x(x+1)}dx < \ln\frac{\pi+2}{2}$$
How prove tasks like this?
 A: hint
use the fact that
$$\forall x\in (0,\frac {\pi}{2}]\;\;\;\frac {2}{\pi}x<\sin (x)<x $$
and integrate to get
$$\frac {2}{\pi}\int_0^{\frac {\pi}{2}}\frac {dx}{1+x}<I<\int_0^{\frac {\pi}{2}} \frac {dx}{1+x} $$
A: We have $\sin x < x$ for $x > 0$ hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x < \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \ln (x+1)\big]_0^{\pi/2} = \ln \left(\frac{\pi}{2} + 1\right).$$
Also $\sin x > \frac{2}{\pi}x$ for $x > 0$ (and $x < \pi/2$) hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x > \frac{2}{\pi} \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \frac{2}{\pi}\ln (x+1)\big]_0^{\pi/2} = \frac{2}{\pi}\ln \left(\frac{\pi}{2} + 1\right).$$
A: The given inequality is a straightforward consequence of the convexity inequality
$$ \forall x\in\left(0,\frac{\pi}{2}\right),\qquad \frac{2}{\pi}x\leq \sin(x)\leq x.\tag{1} $$
The similar convexity inequality
$$ \forall x\in\left(0,\frac{\pi}{2}\right),\qquad 1-\frac{x^2}{6}\leq \frac{\sin(x)}{x}\leq 1-\frac{4(\pi-2)x^2}{\pi^3}\tag{2} $$
leads to much tighter bounds.
