Question

Prove that:
${\int }_0^1{\cot }^{-1}|1-x+x^2|dx$

Answer 1

$I={\int }_0^1{\cot }^{-1}(1-x+x^2)dx$

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

$={\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x(1-x)}\right)dx$

$={\int }_0^1{\tan }^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right)dx$

$={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}(1-x)dx$

$={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}(1-(1-x))dx$ using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}xdx$

$=2{\int }_0^1{\tan }^{-1}xdx$

$=2{\int }_0^1{\tan }^{-1}x.1.dx$

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

$=2[1{\tan }^{-1}1-0]-2\times \frac{1}{2}{\left[\log (1+x^2)\right]}_0^1$

$=2\times \frac{\pi }{4}-(\log 2-\log 1)$

$=\frac{\pi }{2}-\log 2$