JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 1)
The value of the integral $$\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $$ is :-
$${\pi \over 2} - {1 \over 2}{\log _e}2$$
$${\pi \over 4} - {\log _e}2$$
$${\pi \over 4} - {1 \over 2}{\log _e}2$$
$${\pi \over 2} - {\log _e}2$$
Explanation
I = $$\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $$
Let x2 = t
$$ \Rightarrow $$ 2xdx = dt
At x = 0, t = 0
At x = 1, t = 1
I = $${1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t\left( {1 - t} \right)}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
[ As $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$ ]
= $$2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
= $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
Using integration by parts rule
= $$\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1$$$$ - \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt$$
= $${\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1$$
= $${\pi \over 4} - {1 \over 2}{\log _e}2$$
Let x2 = t
$$ \Rightarrow $$ 2xdx = dt
At x = 0, t = 0
At x = 1, t = 1
I = $${1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t\left( {1 - t} \right)}}} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
[ As $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt} $$
= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$ ]
= $$2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
= $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
Using integration by parts rule
= $$\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1$$$$ - \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt$$
= $${\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1$$
= $${\pi \over 4} - {1 \over 2}{\log _e}2$$
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