Question

Evaluate the integral
${\int }_0^1\frac{dx}{\sqrt{1-x^2}}$

• A. $0$
• B. $-1$
• C. $\pi /2$
• D. $-\pi /2$

Answer 1

The correct option is C $\pi /2$

${\int }_0^1\frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+c$

${\int }_0^1\frac{dx}{\sqrt{1-x^2}}={\left[\right({\sin }^{-1}x+c\left)\right]}_0^1$

$=({\sin }^{-1}1+c)-({\sin }^{-1}0+c)$

$=\frac{\pi }{2}$

${\int }_0^1\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$