QJEEM18Mac45c: JEE Main 2018 Mathematics - Class 12 JEEM18Mac45c Solution: If is a constant of integration) then the ordered pair (K, A)…

Question

Verified Solution • 2025–26 Session moderate

If

\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}

\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,

is a constant of integration) then the ordered pair (K, A) is equal to :

Accepted Answer

Given,

\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx

Let tanx = t

\Rightarrow

\,\,\,

sec²x dx = dt

\Rightarrow

\,\,\,

dx =

{{dt} \over {{{\sec }^2}x}}

\Rightarrow

\,\,\,

dx = {{dt} \over {1 + {{\tan }^2}x}}

\Rightarrow

\,\,\,

dx =

{{dt} \over {1 + {t^2}}}

\therefore\,\,\,

\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx

=

\int {{t \over {1 + t + {t^2}}}} \times {{dt} \over {1 + {t^2}}}

=

\int {{t \over {\left( {1 + t + {t^2}} \right)\left( {1 + {t^2}} \right)}}} \,\,dt

=

\int {\left( {{1 \over {1 + {t^2}}} - {1 \over {1 + t + {t^2}}}} \right)} \,\,dt

=

{\tan ^{ - 1}}\left( t \right) - \int {{1 \over {1 + t + {t^2} + {1 \over 4} - {1 \over 4}}}} \,\,dt

=

x - \int {{1 \over {{t^2} + t + {1 \over 4} + 1 - {1 \over 4}}}} \,\,dt

=

x - \int {{1 \over {{{\left( {t + {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}}

=

x - {1 \over {{{\sqrt 3 } \over 2}}} + {\tan ^{ - 1}}\left( {{{2t + 1} \over {\sqrt 3 }}} \right) + c

=

x - {2 \over {\sqrt 3 }}\,\,{\tan ^{ - 1}}\left( {{{2\tan x + 1} \over {\sqrt 3 }}} \right) + c

\therefore\,\,\,

By comparing

A = 3 and K = 2

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