Verified Solution • 2025–26 Session moderate

QJEEM19Ma5c4f: JEE Main 2019 Mathematics - Class 12 JEEM19Ma5c4f Solution: The integral     equals :

The integral   ∫π/6π/4dxsin⁡2x(tan⁡5x+cot⁡5x)\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}   equals :

Multiple Choice Options

  • A π40{\pi \over {40}}
  • B 120tan⁡−1(193){1 \over {20}}{\tan ^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)
  • C 110(π4−tan⁡−1(193)){1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)} \right)
  • D 15(π4−tan⁡−1(133)){1 \over 5}\left( {{\pi \over 4}{{-\tan }^{ - 1}}\left( {{1 \over {3\sqrt 3 }}} \right)} \right)

Step-by-Step Verified Solution

I == ∫π/6π/4dxsin⁡2x(tan⁡5x+cot⁡5x)\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}

I=12∫π/6π/4tan⁡4xsec⁡2xdx(1+tan⁡10x){\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}}   Put tan5x == t

I=110∫(13)51dt1+t2=110(π4−tan⁡−1193){\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)

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