QJEEM22Ma70e8: JEE Main 2022 Mathematics - Class 12 JEEM22Ma70e8 Solution

$= \int {{{\left( {1 - {1 \over {\sqrt 3 }}} \right)(\cos x - \sin x)} \over {\left( {1 + {2 \over {\sqrt 3 }}\sin 2x} \right)}}dx}$

$= \int {{{\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right)\sqrt 2 \sin \left( {{\pi \over 4} - x} \right)} \over {\left( {{2 \over {\sqrt 3 }}} \right)\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx}$

$= \int {{{{{(\sqrt 3 - 1)} \over {\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx}$

$= \int {{{{{\sqrt 3 - 1} \over {2\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}$

$= {1 \over 2}\int {{{2\sin {\pi \over {12}}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}$

$= {1 \over 2}\int {{{\cos \left( {{\pi \over 6} - x} \right) - \cos \left( {{\pi \over 3} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}$

$= {1 \over 2}\left[ {\int {{\mathop{\rm cosec}\nolimits} \left( {{\pi \over 6} + x} \right)dx - \int {\sec \left( {{\pi \over 6} - x} \right)dx} } } \right]$

$= {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \int {\cos ec\left( {{\pi \over 3} - x} \right)dx} } \right]$

$= {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \ln \left| {{\pi \over 6} + {x \over 2}} \right|} \right] + C$

$= {1 \over 2}\ln \left| {{{\tan \left( {{\pi \over 2} + {x \over 2}} \right)} \over {\tan \left( {{\pi \over 6} + {x \over 2}} \right)}}} \right| + C$