If [ x 1 + x + ^2xdx = x - K A ^ - 1 ] [( K, x + 1 A ) + C,(C,,] is a constant of integrat…

Question

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

Solutions · 1