Question

Let $C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(nx\right)}{{\sin }^{-1}\left(nx\right)}dx.$ Then $\underset{n\to \infty }{lim}{n}^2.C_n$ equals

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

$C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}\left(nx\right)}{{\sin }^{-1}\left(nx\right)}dx$
Put $nx=t\Rightarrow dx=\frac{dt}{n}$
$C_n=\frac{1}{n}\underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt$

Now, $L=\underset{n\to \infty }{lim}{n}^2\cdot C_n$
$=\underset{n\to \infty }{lim}n\underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt(\infty \times 0)\text{form}$
$L=\underset{n\to \infty }{lim}\frac{\underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt}{1/n}\left(\frac{0}{0}\right)\text{form}$
Applying Leibnitz rule,
$L=\underset{n\to \infty }{lim}\frac{0-\frac{{\tan }^{-1}\frac{n}{n+1}}{{\sin }^{-1}\frac{n}{n+1}}\left(\frac{1}{(n+1{)}^2}\right)}{-\frac{1}{n^2}}$
$=\frac{\pi }{4}\times \frac{2}{\pi }=\frac{1}{2}$