Question

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

• A. $\frac{{\pi }^3}{24}$
• B. ${\pi }^2$
• C. $-{\pi }^2$
• D. $0$

Answer 1

The correct option is A $\frac{{\pi }^3}{24}$

${\int }_0^1\frac{(si{n}^{-1}x{)}^2}{\sqrt{1-x^2}}\cdot dx$

$\int \frac{(si{n}^{-1}x{)}^2}{\sqrt{1-x^2}}\cdot dx=\int \left(si{n}^{-1}x\right)d\left(si{n}^{-1}x\right)$

$=\frac{(si{n}^{-1}x{)}^3}{3}+c$

${\int }_0^1\frac{(si{n}^{-1}x{)}^2}{\sqrt{1-x^2}}\cdot dx={[\frac{(si{n}^{-1}x{)}^3}{3}+c]}_0^1$

$=(\frac{(si{n}^{-1}1{)}^3}{3}+c)-(\frac{(si{n}^{-1}0{)}^3}{3}+c)$

$=\frac{1}{3}{\left(\frac{\pi }{2}\right)}^3$

$=\frac{{\pi }^3}{24}$