Question Number 115362 by Bird last updated on 25/Sep/20
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Answered by TANMAY PANACEA last updated on 25/Sep/20
$$\mathrm{1}\geqslant{sinx}\geqslant−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\geqslant\frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} }\geqslant\frac{−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${tan}^{−\mathrm{1}} {x}\mid_{\mathrm{0}} ^{\mathrm{1}} \geqslant{I}\geqslant−{tan}^{−\mathrm{1}} {x}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\frac{\pi}{\mathrm{4}}\geqslant{I}\geqslant−\frac{\pi}{\mathrm{4}} \\ $$