Question Number 143904 by nadovic last updated on 19/Jun/21
Answered by Dwaipayan Shikari last updated on 19/Jun/21
$$\frac{\mathrm{1}}{\:\sqrt{\pi}}\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\mathrm{1}\:\:{e}^{\pi{i}} =−\mathrm{1} \\ $$$${p}=\mathrm{2} \\ $$$$\frac{{d}}{{dx}}\mid_{{x}=\frac{\mathrm{1}}{\mathrm{4}}} \left({xsin}^{−\mathrm{1}} {px}+\frac{\sqrt{\mathrm{1}−{p}^{\mathrm{2}} {x}^{\mathrm{2}} }}{{p}}\right) \\ $$$$=\frac{{xp}}{\:\sqrt{\mathrm{1}−{p}^{\mathrm{2}} {x}^{\mathrm{2}} }}+{sin}^{−\mathrm{1}} {px}−\frac{{xp}}{\:\sqrt{\mathrm{1}−{p}^{\mathrm{2}} {x}^{\mathrm{2}} }}={sin}^{−\mathrm{1}} {px} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:{it}\:{is}\:{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{6}} \\ $$