Question

Evaluate the integral ${\int }_0^{1}co{s}^{-1}xdx$

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

Answer 1

The correct option is A $1$

$I={\int }_0^1{\cos }^{-1}xdx={\left[x{\cos }^{-1}x\right]}_0^1-{\int }_0^{1}x-\frac{1}{\sqrt{1-x^2}}dx$
$=0+{\int }_0^1\frac{x}{\sqrt{1-x^2}}dx$
Substitute $1-x^2=t\Rightarrow -2xdx=dt\Rightarrow xdx=-\frac{1}{2}$
$\therefore I=-\frac{1}{2}{\int }_1^0\frac{1}{\sqrt{t}}dt$
$\frac{1}{2}{\int }_0^1\frac{1}{\sqrt{t}}dt{\left[\frac{1}{2}\frac{\sqrt{t}}{\frac{1}{2}}\right]}_0^0=1$