Question

$\underset{0}{\overset{1}{\int }}{\cos }^2{\sin }^{-1}xdx$ is equal to

• A. $1$
• B. $\frac{4}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

$\underset{0}{\overset{1}{\int }}{\cos }^2{\cos }^{-1}\sqrt{1-x^2}dx=\underset{0}{\overset{1}{\int }}{\left(\cos {\cos }^{-1}\sqrt{1-x^2}\right)}^{2}dx=\underset{0}{\overset{1}{\int }}(\sqrt{1-x^2}{)}^{2}dx$
$=\underset{0}{\overset{1}{\int }}(1-x^2)dx=[x-\frac{x^3}{3}{]}_0^1=1-\frac{1}{3}=\frac{2}{3}$