Question

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

• A. $\pi -\frac{\log }{2}$
• B. $2\pi -\frac{\log }{2}$
• C. $\pi +\frac{\log }{2}$
• D. None of these

Answer 1

The correct option is A $\pi -\frac{\log }{2}$

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

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

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

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

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

Taking dummy valuable approach :

Let $(1-x)=t$

$-dx=dt$

$x\to 0$ to $1$

$\Rightarrow t\to 1$ to $0$

$={\int }_0^1{\tan }^{-1}xdx-{\int }_1^0{\tan }^{-1}dt$

$=2{\int }_0^1{\tan }^{-1}xdx(Replacingdummy$ $\because {\int }_a^{b}f\left(t\right)dt={\int }_b^{a}b\left(t\right)dt$

$=2(\frac{\pi }{4}-\frac{\log e^2}{2})$

$=\frac{\pi }{4}-{\log }^2$

$\stackrel{\uparrow }{{\int }_0^1{\tan }^{-1}xdx}$

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

By part integration

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

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

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

$\frac{\pi }{4}-\frac{\log 2}{2}$

$=\frac{\pi }{4}-{\log }^2$