Let $C_n =\int^{1/n}_{1/(n+1)}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx $ 
Let
$$C_n =\int^{1/n}_{1/(n+1)}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx\textrm{.} $$
Then $\lim_{n \to \infty} n ^2C_n$ equals :

(a) $1$
(b) $0$
(c) $-1$
(d) $\frac12$
I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.
 A: You can use the Mean Value Theorem for integrals to get the limit. In fact, since
$$ \int_{\frac{1}{n+1}}^{\frac1n}\frac{\arctan nx}{\arcsin nx}dx=\frac{\arctan n\xi_n}{\arcsin n\xi_n}\frac{1}{n(n+1)}$$
where 
$$\frac{1}{n+1}<\xi_n<{\frac1n},$$
we have
\begin{eqnarray}
\lim_{n\to\infty}n^2C_n&=&\lim_{n\to\infty}n^2\int_{\frac{1}{n+1}}^{\frac1n}\frac{\arctan nx}{\arcsin nx}dx\\
&=&\lim_{n\to\infty}n^2\frac{\arctan n\xi_n}{\arcsin n\xi_n}\frac{1}{n(n+1)}\\
&=&\lim_{n\to\infty}\frac{\arctan n\xi_n}{\arcsin n\xi_n}.
\end{eqnarray}
Since 
$$\frac{1}{n+1}<\xi_n<{\frac1n}$$
we have
$$\frac{n}{n+1}<n\xi_n<1$$
or $\lim_{n\to\infty}n\xi_n=1$. So
$$ \lim_{n\to\infty}n^2C_n=\frac{\arctan 1}{\arcsin 1}=\frac{\frac{\pi}{4}}{\frac{\pi}{2}}=\frac12.$$
So you have to choose (d).
A: the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration. 
it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=\frac{dz}{n}$. this gives altogether:
$$
C_n = \frac1n \int_{\frac{n}{n+1}}^1 \frac {\tan^{-1} z}{\sin^{-1}z} dz
$$
you should be able to proceed from here, using the asymptotic equality of the ratio of $\tan z$ and $\sin z$ as $z \to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt
