Question Number 100948 by Dwaipayan Shikari last updated on 29/Jun/20
Commented by Dwaipayan Shikari last updated on 29/Jun/20
$$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{r}}{{n}}\right){sec}^{\mathrm{2}} \left(\frac{{r}}{{n}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {xsec}^{\mathrm{2}} {x}^{\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xsec}^{\mathrm{2}} {x}^{\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {sec}^{\mathrm{2}} {tdt} \\ $$$${Suppose}\:{x}^{\mathrm{2}} ={t}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{tan}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{tan}\mathrm{1} \\ $$$$ \\ $$