Question

$\stackrel{Lt}{x\to 0}{\int }_0^x\frac{(Ta{n}^{-1}t{)}^2}{\sqrt{1+x^2}}dt=$

• A. $0$
• B. $\frac{{\pi }^2}{4}$
• C. ${\pi }^2$
• D. $\frac{{\pi }^2}{2}$

Answer 1

The correct option is A $0$

By L - Hospital's Rule
$=\stackrel{Lt}{x\to 0}\frac{(ta{n}^{-1}x{)}^2\sqrt{1+x^2}}{x}$
$=\stackrel{Lt}{x\to 0}{\left(\frac{ta{n}^{-1}x}{x}\right)}^{2}x\sqrt{1+x^2}=0$