If $\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\arctan(nx)}{\arcsin(nx)}dx=c_n$ then $\lim_{n\to\infty}{n^2c_n}=?$

Question

$$\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\arctan(nx)}{\arcsin(nx)}dx=c_n$$

Find the value of $$\lim_{n\to\infty}{n^2c_n}$$

I tried integrating the expression but couldn't. Need help.

Answer 1

Make the substitution $u=nx$ to get

$$\int_{\frac{n}{n+1}}^1 \frac{\arctan u}{\arcsin u} \frac{1}{n} \; du.$$

Then, $\arcsin(1) = \pi/2$ and $\arctan(1)=\pi/4$. For large $n$, you're integrating on a tiny interval close to $1$, so your integrand is very close to $1/2$. So you should have

$$c_n \approx \int_{\frac{n}{n+1}}^1 \frac{1}{2n} \; du = \frac{1}{2n(n+1)},\,n^2c_n\approx\frac{n}{2(n+1)},$$

which would make your limit equal to $1/2$.

So you have to argue that "very close" really does cause a good enough $\approx$.