Let \(C_n=\displaystyle\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}\,dx\), then \(\displaystyle\lim_{n\to\infty}n^2\cdot C_n\) equals
$\frac{1}{2}$
Substitute \(y=nx\), so \(dx=dy/n\); the limits become \(y=\dfrac{n}{n+1}\) to \(y=1\).
So \(C_n=\dfrac1n\displaystyle\int_{n/(n+1)}^{1}\dfrac{\tan^{-1}y}{\sin^{-1}y}\,dy\). Let \(f(y)=\dfrac{\tan^{-1}y}{\sin^{-1}y}\), so \(f(1)=\dfrac{\pi/4}{\pi/2}=\dfrac12\).
By the Mean Value Theorem for integrals, \(\displaystyle\int_{n/(n+1)}^1 f(y)\,dy=f(\xi_n)\left(1-\dfrac{n}{n+1}\right)=f(\xi_n)\cdot\dfrac{1}{n+1}\), for some \(\xi_n\) between \(n/(n+1)\) and 1; as \(n\to\infty\), \(\xi_n\to 1\), so \(f(\xi_n)\to \dfrac12\) by continuity of \(f\) at 1.
Hence \(C_n\approx \dfrac{1}{n}\cdot\dfrac{1}{n+1}\cdot f(\xi_n)\), so \(n^2C_n=\dfrac{n}{n+1}f(\xi_n)\).
Taking \(n\to\infty\): \(\dfrac{n}{n+1}\to 1\) and \(f(\xi_n)\to \dfrac12\), so \(\displaystyle\lim_{n\to\infty}n^2C_n=\dfrac12\).
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