Question

If $\int \frac{x\cot x+1}{\sqrt{cose{c}^{2}x-x^2}}dx=f\left(x\right)+c,$ where $\underset{x\to 0^{+}}{lim}f\left(x\right)=0$ and $\text{cosec}x>0$. (where $c$ is a constant of integration), then which of the following is/are incorrect

• A. $f\left(1\right)=1$
• B. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{|x|\tan x}$ does not exist.
• C. $f\left(1\right)=\frac{\pi }{2}-1$
• D. $f\left(x\right)$ is periodic with a period $2\pi$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\underset{x\to 0}{lim}\frac{f\left(x\right)}{|x|\tan x}$ does not exist.
C $f\left(1\right)=\frac{\pi }{2}-1$
D $f\left(x\right)$ is periodic with a period $2\pi$

$I=\int \frac{x\cos x+\sin x}{\sqrt{1-(x\sin x{)}^2}}dx={\sin }^{-1}\left(x\sin x\right)+c$
$f\left(x\right)={\sin }^{-1}\left(x\sin x\right)\Rightarrow f\left(1\right)={\sin }^{-1}\left(\sin 1\right)=1$
(as $\text{cosec}x>0$) so $x\to 0^{+}$
$\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)}{|x|\tan x}=\frac{{\sin }^{-1}\left(x\sin x\right)}{|x|\sin x}\times \cos x=\frac{{\sin }^{-1}\left(x\sin x\right)}{x\sin x}\times \cos x$
$=1$