Let $C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(nx\right)}{{\sin }^{-1}\left(nx\right)}dx$  then  $\underset{n\to \infty }{\lim }{n}^2\cdot C_n$ equals :

$C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(nx\right)}{{\sin }^{-1}\left(nx\right)}dx$  (Put $nx=t$)
 $$\begin{gathered} C_n=\frac{1}{n}\underset{\frac{n}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(t\right)}{{\sin }^{-1}\left(t\right)}dx \\ L=\underset{n\to \infty }{\lim }{n}^2\cdot C_n=\underset{n\to \infty }{\lim }n\cdot \underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt(\infty \times 0) \\ L=\underset{n\to \infty }{\lim }{n}^2\cdot C_n=\underset{n\to \infty }{\lim }n\cdot \underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt(\infty \times 0) \end{gathered}$$
 Applying Leibnitz rule
 $L=\underset{n\to \infty }{\lim }\frac{0-\frac{{\tan }^{-1}\frac{n}{n+1}}{{\sin }^{-1}\frac{n}{n+1}}\left(\frac{1}{{(n+1)}^2}\right)}{-\frac{1}{n^2}}=\frac{\pi }{4}\cdot \frac{2}{\pi }=\frac{1}{2}$