Question

If $x\ge 1$ and $\int \sin \left({\cot }^{-1}\sqrt{x^2-1}\right)dx=f\left(x\right)+k,$ where $f\left(1\right)=0$, then the value of $f\left(e^2\right)$ is equal to
(Here, $k$ is a constant of integration.)

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

Answer 1

The correct option is B $2$

Since ${\cot }^{-1}\sqrt{x^2-1}={\sin }^{-1}\frac{1}{x}$
$\therefore \sin \left({\cot }^{-1}\sqrt{x^2-1}\right)=\sin \left({\sin }^{-1}\frac{1}{x}\right)$
$\Rightarrow \sin \left({\cot }^{-1}\sqrt{x^2-1}\right)=\frac{1}{x}$
Now, $\int \sin \left({\cot }^{-1}\sqrt{x^2-1}\right)dx$
$=\int \frac{1}{x}dx=\ln x+k$
$\therefore f\left(x\right)=\ln x$
$\Rightarrow f\left(e^2\right)=2$