If [2_0^1 ^ - 1xdx = _0^1 ^ - 1 ( 1 - x + x^2 )dx,] then [_0^1 ^ - 1 ( 1 - x + x^2 )dx] is…

Question

If   $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$

then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto :

Accepted Answer

Given,

$$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}}  = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$

=  $$\int\limits_0^1 {\left( {{\pi  \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$$

$$ \Rightarrow $$ $\,\,\,$$ $2\int\limits\_0\^1 {{{\tan }\^{ - 1}}} x\,dx = \int\limits\_0\^1 {{\pi  \over 2}} dx - \int\limits\_0\^1 {{{\tan }\^{ - 1}}} \left( {1 - x + {x\^2}} \right)dx$$

$$ \Rightarrow $$ $\,\,\,$$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi  \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx} $$ \Rightarrow $$ $\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi  \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} $$

Let  I = $$\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx$$

=   $$\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx$$

=  $${\pi  \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1$$

=  $${\pi  \over 4} - {1 \over 2}\log 2$$

$$\therefore\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx$$

=  $${\pi  \over 2} - 2\left( {{\pi  \over 4} - {1 \over 2}\log 2} \right)$$

=  $${\pi  \over 2} - {\pi  \over 2} + \log 2$$

=   log2

Solutions · 1