Question 142 of 246MediumMCQ
The value of the integral \[\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} \] is :-
Solutions · 1
Official solution
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
= \[\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1\] −0∫1t.1+t21dt=4π−21[log(1+t2)]01=4π−21loge2
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