
man111 singh Nice Solution hsbhatt Sir.Upvote·0· Reply ·2013-10-01 00:49:35
Limn→ ∞ [ n2 . ∫ ( tan-1 nx)/( sin-1 nx) dx
Integration upper limit:- ( 1/n )
Integration lower limit:- { (1+ n ) / n }

Newton-Leibniz might help
\hspace{-16}$ Is yours question is $\bf{\lim_{n\rightarrow \infty}n^2 \cdot \int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx}$\\\\\\ Then Answer is $\bf{=\frac{1}{2}}$