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Question Number 100948 by Dwaipayan Shikari last updated on 29/Jun/20

Commented by Dwaipayan Shikari last updated on 29/Jun/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{r=1}{\sum ^n}\left(\frac{r}{n}\right){sec}^2{\left(\frac{r}{n}\right)}^2$$$${\int }_0^1{xsec}^2{x}^{2}dx=\frac{1}{2}{\int }_0^{1}2{xsec}^2{x}^{2}dx=\frac{1}{2}{\int }_0^1{sec}^{2}tdt$$$$Suppose{x}^2=t=\frac{1}{2}{\left[tant\right]}_0^1=\frac{1}{2}tan1$$$$$$

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