QJEEM16Ma6a92: JEE Main 2016 Mathematics - Class 12 JEEM16Ma6a92 Solution: If   then is equalto :

Question

Verified Solution • 2025–26 Session moderate

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

QJEEM16Ma6a92: JEE Main 2016 Mathematics - Class 12 JEEM16Ma6a92 Solution: If&nbsp;&nbsp; then is equalto :