$\lim_{n\rightarrow\infty} n^2C_n$ for $C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$ equals? Let $$C_n=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\tan^{-1}nx}{\sin^{-1}nx}dx$$ then $\lim_{n\rightarrow\infty} n^2C_n$ equals?
I am having trouble in finding the integral.Wolfram alpha too doesnt give any answer http://www.wolframalpha.com/input/?i=integrate+arctan(nx)%2Farcsin(nx).
Any idea?
 A: try a substitution $x n= q$, then
$$
C_n=\frac{1}{n}\int_{\frac{n}{n+1}}^1\frac{\arctan(q)}{\arcsin(q)}dq
$$
since $\frac{n}{n+1}\sim1 $ for $n \rightarrow \infty$ we might expand the integrand in a taylorseries around $q=1$ which means $\arctan(x)\sim \frac{\pi}{4}$ and $\arcsin(x)\sim \frac{\pi}{2}$
$$
C_n\sim\frac{1}{2n}\int_\frac{n}{n+1}^1dq\ \sim\frac{1}{2 n^2}
$$
therefor 

$$
\lim_{n\rightarrow \infty}n^2 C_n=\frac{1}{2}
$$

A: Making the transformation $x \mapsto x/n$, we see $$C_n = n \int^1_{\tfrac n {n+1}} \frac{\arctan x}{ \arcsin x} dx.$$ From here, put $$f(t) = \int_t^1 \frac{\arctan x}{ \arcsin x} dx.$$ Expanding in a Taylor series, we see that for $t$ near $1$, we have $$f(t) = f(1) + (t-1)f'(1) + \mathcal O((t-1)^2).$$ But $f(1) = 0$ and $$f'(t)= - \frac{\arctan(t)}{\arcsin(t)} \,\,\, \implies \,\,\, f'(1) = -\frac{\pi/4}{\pi / 2} = -\frac 1 2$$ so for very large $n$, we have $$C_n = n f\left( \tfrac n {n+1} \right) = n \cdot \tfrac 1 2 \left(1-\tfrac n {n+1} \right) + n \mathcal O \left( \left( 1-\tfrac n {n+1} \right)^2 \right). $$ Finally, $1 - \tfrac n {n+1} = \tfrac 1 {n+1}$ so we get $$C_n = \frac 1 2\frac n {n+1} + \mathcal O \left( \frac 1 n\right) $$ so $$\lim_{n\to\infty} C_n = \frac 1 2. $$
A: By the MVT for integrals, there is $\xi_n\in[\frac{1}{n+1},\frac1n]$ such that
$$ \int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx=\frac{\tan^{-1}(n\xi_n)}{\sin^{-1}(n\xi_n)}\frac{1}{n(n+1)}.$$
Note that both $\sin^{-1}x$ and $\tan^{-1}x$ are increasing in $[0,1]$ and hence
$$\frac{\tan^{-1}(n\cdot\frac1{n+1})}{\sin^{-1}(n\cdot\frac1{n})}\le\frac{\tan^{-1}(n\xi_n)}{\sin^{-1}(n\xi_n)}\le \frac{\tan^{-1}(n\cdot\frac1n)}{\sin^{-1}(n\cdot\frac1{n+1})}. $$
So
$$ n^2\frac{\tan^{-1}(n\cdot\frac1{n+1})}{\sin^{-1}(n\cdot\frac1{n})}\frac{1}{n(n+1)}\le n^2\int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx\le n^2\frac{\tan^{-1}(n\cdot\frac1n)}{\sin^{-1}(n\cdot\frac1{n+1})}\frac{1}{n(n+1)} $$
which implies
$$ \lim_{n\to\infty}n^2\int_\frac{1}{n+1}^\frac1n\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx=\frac12.$$
A: My idea is to split the integral into $\int_0^{1/n} - \int_0^{1/(n+1)}$.  Then put the $n^2$ in the denominator as $1/n^2$.  Now you should be set up for L'hospital's rule:
$$\lim_{n\rightarrow \infty} n^2C_n = \lim_{n\rightarrow \infty}\frac{\int_0^{1/n} \frac{\tan^{-1}nx}{\sin^{-1}nx} \;dx - \int_0^{1/(n+1)} \frac{\tan^{-1}nx}{\sin^{-1}nx}\; dx}{\frac{1}{n^2}}$$
Take the derivative of top and bottom:
$$=\lim_{n\rightarrow \infty} \frac{ \frac{\tan^{-1}1}{\sin^{-1}1} - \frac{\tan^{-1}\frac{n+1}{n}}{\sin^{-1}\frac{n+1}{n}}}{\frac{-2}{n^3}}$$
One more application of L'hospital should (?) give a mostly algebraic expression.
