Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

$=2[\frac{\pi }{4}-0]-{\left[\log (1+x^2)\right]}_0^1$

$=\frac{\pi }{2}-[\log 2-\log 1]$

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