Question

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

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

Answer 1

Answer: (B)

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