Question

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

• A. $\pi$
• B. $\pi /2$
• C. $\pi /3$
• D. $\pi /4$

Answer 1

Answer: (C)

$\pi /3$

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

We know that $\int \frac{1}{\sqrt{1-x^2}}dx=({\sin }^{-1}x+c)$

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

$=(si{n}^{-1}1+c)-(\sin \frac{1}{2}+c)$

$={\sin }^{-1}1-{\sin }^{-1}\frac{1}{2}$

$=\frac{\pi }{2}-\frac{\pi }{6}$

$=\frac{\pi }{3}$

${\int }_{\frac{1}{2}}^1\frac{1}{\sqrt{1-x^2}}dx=\frac{\pi }{3}$