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QJEEM19Ma524a: JEE Main 2019 Mathematics - Class 12 JEEM19Ma524a Solution: The value of the integral is :-

The value of the integral ∫01xcot⁡−1(1−x2+x4)dx\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} is :-

Multiple Choice Options

  • A π2−12log⁡e2{\pi \over 2} - {1 \over 2}{\log _e}2
  • B π4−log⁡e2{\pi \over 4} - {\log _e}2
  • C π4−12log⁡e2{\pi \over 4} - {1 \over 2}{\log _e}2
  • D π2−log⁡e2{\pi \over 2} - {\log _e}2

Step-by-Step Verified Solution

I = ∫01xcot⁡−1(1−x2+x4)dx\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 = 12∫01cot⁡−1(1−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt}

= 12∫01tan⁡−1(11−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt}

= 12∫01tan⁡−1(t+(1−t)1−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt}

= 12∫01tan⁡−1(t+(1−t)1−t(1−t))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}

= 12∫01[tan⁡−1(1−t)+tan⁡−1(t)]dt{1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt}

= 12∫01tan⁡−1(t)dt+12∫01tan⁡−1(1−t)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}

= 12∫01tan⁡−1(t)dt+12∫01tan⁡−1(t)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 12∫01tan⁡−1(1−t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt}

= 12∫01tan⁡−1[0+1−(1−t)]dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt}

= 12∫01tan⁡−1(t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} ]

= 2×12∫01tan⁡−1(t)dt2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

= ∫01tan⁡−1(t)dt\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

Using integration by parts rule

= [t.tan⁡−1(t)]01\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1 −∫01t.11+t2dt- \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt

= π4−12[log⁡(1+t2)]01{\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1

= π4−12log⁡e2{\pi \over 4} - {1 \over 2}{\log _e}2

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