Question

If \[2\int_0^1 {{{\tan }^{ - 1}}xdx} = \int_0^1 {{{\cot }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} \],then $\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} $ is equal to-
A)$\log 2$ B)$\dfrac{\pi }{2} - \log 4$ C)$\dfrac{\pi }{2} + \log 2$ D)$\log 4$

Answer 1

Hint- We can use trigonometric identity ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ to find$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} $ and then normally integrate the obtained equation.

Complete step-by-step answer:
Given $2\int_0^1 {{{\tan }^{ - 1}}xdx} = \int_0^1 {{{\cot }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} $ and we know that${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ , and then we can rewrite the given equation as
$2\int_0^1 {{{\tan }^{ - 1}}xdx} = \int_0^1 {\dfrac{\pi }{2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} $
Now let$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx} $=I=$\dfrac{\pi }{2} - 2\int_0^1 {{{\tan }^{ - 1}}xdx} $
Now on integrating $\int_0^1 {{{\tan }^{ - 1}}xdx} $,we have
   $
  \dfrac{\pi }{2} - 2\left\{ {\left( {{{\tan }^{ - 1}}x\int {1dx} } \right)_0^1 - \left[ {\left( {\dfrac{{d{{\tan }^{ - 1}}x}}{{dx}}} \right)\int {1dx} } \right]_0^1} \right\} \\
    \\
 $
I=$\dfrac{\pi }{2} - 2\left[ {{{\tan }^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - \dfrac{1}{2}\left\{ {\log \left( {1 + {x^2}} \right)} \right\}_0^1} \right]$
I=$\dfrac{\pi }{2} - 2\left[ {\dfrac{\pi }{4} - \dfrac{1}{2}\log 2} \right]$
I=$\dfrac{\pi }{2} - \dfrac{\pi }{2} + \log 2$ =$\log 2$
Hence the correct answer is A.

Note: Here, we have solved$\int_0^1 {{{\tan }^{ - 1}}xdx} $ by chain rule and to solve$\int {\dfrac{x}{{\left( {1 + {x^2}} \right)}}} dx$ put$\left( {1 + {x^2}} \right) = t \Rightarrow xdx = \dfrac{{dt}}{2}$.On putting this value the integration becomes simple and you can easily get the answer.