JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 6)
The integral $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$ equals :
$${\pi \over {40}}$$
$${1 \over {20}}{\tan ^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)$$
$${1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)} \right)$$
$${1 \over 5}\left( {{\pi \over 4}{{-\tan }^{ - 1}}\left( {{1 \over {3\sqrt 3 }}} \right)} \right)$$
Explanation
I $$=$$ $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$
$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$ Put tan5x $$=$$ t
$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$
$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$ Put tan5x $$=$$ t
$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$
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