Question

If ${\int }_0^1{\cot }^{-1}(1-x+x^2)dx=\lambda {\int }_0^1{\tan }^{-1}xdx$, then $\lambda$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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

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

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

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
We get,
$I={\int }_0^1\left({\tan }^{-1}x\right)dx+{\int }_0^1\left({\tan }^{-1}(1-(1-x))\right)dx$

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

Which gives, $\lambda =2$

Hence, option B.