Question

If $\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{{\cot }^{-1}(1-x+x^2)}dx$ equals $k,$ then the value of $\frac{1}{k}$ is

• A. 2.00
• B. 02
• C. 2
• D. 2.0

Answer 1

Answer: (A), (B), (C), (D)

$k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{{\cot }^{-1}(1-x+x^2)}dx\dots \left(1\right)\Rightarrow k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{{\cot }^{-1}\left(\right(1-x{)}^2+x)}dx$
Applying $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx,$
$k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}(1-x)}{{\cot }^{-1}\left(\right(1-1+x{)}^2+(1-x))}dx\Rightarrow k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}(1-x)}{{\cot }^{-1}(x^2+1-x)}dx\dots \left(2\right)$

From $\left(1\right)+\left(2\right)$
$2k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}(1-x)+{\tan }^{-1}\left(x\right)}{{\cot }^{-1}(x^2+1-x)}dx\Rightarrow 2k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}\left(\frac{x+1-x}{1+x(1-x)}\right)}{{\cot }^{-1}(x^2+1-x)}dx\Rightarrow 2k=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}\left(\frac{1}{x^2+1-x}\right)}{{\cot }^{-1}(x^2+1-x)}dx\Rightarrow 2k=\underset{0}{\overset{1}{\int }}1dx\Rightarrow k=\frac{1}{2}$
$\therefore \frac{1}{k}=2$