JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 2)
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 :
(2, 1)
($$-$$2, 3)
(2, 3)
($$-$$2, 1)
Explanation
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
$$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$$
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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