Question

The value of ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}co{s}^{-1}xdx$ is

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

Answer 1

Answer: (B)

$\underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}{\cos }^{-1}xdx$
$={\left|x{\cos }^{-1}x\right|}_{-\frac{1}{2}}^{\frac{1}{2}}+\underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}\frac{x}{\sqrt{1-x^2}}dx$ [Using integration by parts]
$=\frac{1}{2}{\cos }^{-1}\left(\frac{1}{2}\right)+\frac{1}{2}{\cos }^{-1}\left(-\frac{1}{2}\right)-{\left[\left(\frac{1}{2}\right)\cdot 2\sqrt{1-x^2}\right]}_{-\frac{1}{2}}^{\frac{1}{2}}$
$=\frac{1}{2}\cdot \frac{\pi }{3}+\frac{1}{2}\cdot \frac{2\pi }{3}-0$
$=\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$