<s>[INST] 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?
 [/INST] 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 $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x > \frac{2}{\pi} \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \frac{2}{\pi}\ln (x+1)\big]_0^{\pi/2} = \frac{2}{\pi}\ln \left(\frac{\pi}{2} + 1\right).$$
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