Question

If $I={\int }_0^1{\cot }^{-1}(1-x+x^2)dx=k{\int }_0^1{\tan }^{-1}xdx,$ then $k$ equals

• A. $1$
• B. $2$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is B $2$

$I={\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x(1-x)}\right)dx$
$={\int }_0^1{\tan }^{-1}\left(\frac{(1-x)+x}{1-(1-x)x}\right)dx$
$={\int }_0^1\left[{\tan }^{-1}\right(1-x)+{\tan }^{-1}x]dx$
$={\int }_0^1{\tan }^{-1}(1-x)dx+{\int }_0^1{\tan }^{-1}xdx$
$={\int }_0^1{\tan }^{-1}(1-(1-x\left)\right)dx+{\int }_0^1{\tan }^{-1}xdx$
$=2{\int }_0^1{\tan }^{-1}xdx$