Question

If $A={\int }_0^1{\sin }^{-1}xdx$ and $B={\int }_0^1{\cos }^{-1}xdx$ then $\frac{\pi }{2}$ is equal

• A. $A-B+2$
• B. $B-2B$
• C. $A+1$
• D. $A+B$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $A-B+2$
C $A+1$
D $A+B$

$A={\int }_0^1{\sin }^{-1}xdx={\left[x{\sin }^{-1}x\right]}_0^1-{\int }_0^1\frac{x}{\sqrt{1-x^2}}dx$
$=\frac{\pi }{2}-0+{\left[\sqrt{1-x^2}\right]}_0^1=\frac{\pi }{2}-1$
$B={\int }_0^1{\cos }^{-1}xdx={\left[x{\cos }^{-2}x\right]}_0^1+{\int }_0^1\frac{x}{\sqrt{1-x^2}}dx=0+{[-\sqrt{1-x^2}]}_0^1=+1$