Question

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

• A. 1
• B. 2
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is B 2

We have,
$I={\int }_0^1{\cot }^{-1}(1-x+x^2)\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}\left(\frac{1}{(1-x+x^2)}\right\}\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x(1-x)}\right\}\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}\left(\frac{\left(1-x\right)+x}{1-x(1-x)}\right\}\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1\left({\tan }^{-1}(1-x)+{\tan }^{-1}x\right\}\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}(1-x)\mathrm{dx}+{\int }_0^1{\tan }^{-1}x\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}\left(1-(1-x)\right]\mathrm{dx}+{\int }_0^1{\tan }^{-1}x\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow {\int }_0^1{\tan }^{-1}x\mathrm{dx}+{\int }_0^1{\tan }^{-1}x\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow 2{\int }_0^1{\tan }^{-1}x\mathrm{dx}=k{\int }_0^1{\tan }^{-1}x\mathrm{dx}\Rightarrow k=2$