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Q.
Let $C_n=\int\limits_{\frac{1}{n+1}}^{\frac{1}{n}} \frac{\tan ^{-1}(n x)}{\sin ^{-1}(n x)} d x$ then $\operatorname{Lim}_{n \rightarrow \infty} n^2 \cdot C_n$ equals
Integrals
Solution:
$C_n=\int\limits_{\frac{1}{n+1}}^{\frac{1}{n}} \frac{\tan ^{-1}(n x)}{\sin ^{-1}(n x)} d x \quad$ (put nx $\left.=t\right) \Rightarrow \quad C_n=\frac{1}{n} \int\limits_{\frac{n}{n+1}}^1 \frac{\tan ^{-1}(t)}{\sin ^{-1}(t)} d t$
$L =\underset{ n \rightarrow \infty}{\text{Lim}} n^2 \cdot C _{ n }=\underset{ n \rightarrow \infty}{\text{Lim}} n \cdot \int\limits_{\frac{ n }{ n +1}}^1 \frac{\tan ^{-1} t}{\sin ^{-1} t} dt \quad(\infty \times 0) ; \quad L =\frac{\frac{\tan ^{-1} t}{\sin ^{-1} t} d t}{\frac{1}{n}} \quad\left(\frac{0}{0}\right)$
applying Leibnitz rule
$L =\underset{ n \rightarrow \infty}{\text{Lim}} \frac{0-\frac{\tan ^{-1} \frac{ n }{ n +1}}{\sin ^{-1} \frac{ n }{ n +1}}\left(\frac{1}{( n +1)^2}\right)}{-\frac{1}{ n ^2}}=\frac{\pi}{4} \cdot \frac{2}{\pi}=\frac{1}{2}$