QJEEM22Ma885a: JEE Main 2022 Mathematics - Class 12 JEEM22Ma885a Solution: For , if , then

Given,

$I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx}$

$= \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }$

$= \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }$

$= {1 \over {{{\sin }^{2022}}x}}\,.\,\tan x - \int {\left( {{{ - 2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,\tan x} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} }$

$= {{\tan x} \over {{{\sin }^{2022}}x}} + \int {\left( {{{2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,{{\sin x} \over {\cos x}}} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} }$

$= {{\tan x} \over {{{\sin }^{2022}}x}} + \int {{{2022} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }$

$= {{\tan x} \over {{{\sin }^{2022}}x}} + C$

Given, $I\left( {{\pi \over 4}} \right) = {2^{1011}}$

$\therefore$ $I\left( {{\pi \over 4}} \right) = {{\tan \left( {{\pi \over 4}} \right)} \over {{{\left( {\sin {\pi \over 4}} \right)}^{2022}}}} + C$

$\Rightarrow {2^{1011}} = {1 \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^{2022}}}} + C$

$\Rightarrow C = {2^{1011}} - {2^{1011}} = 0$

$\therefore$ $I(x) = {{\tan x} \over {{{\sin }^{2022}}x}}$

$\therefore$ $I\left( {{\pi \over 3}} \right) = {{\tan {\pi \over 3}} \over {{{\left( {\sin {\pi \over 3}} \right)}^{2022}}}} = {{\sqrt 3 } \over {{{\left( {{{\sqrt 3 } \over 2}} \right)}^{2022}}}}$

$I\left( {{\pi \over 6}} \right) = {{{1 \over {\sqrt 3 }}} \over {{{\left( {{1 \over 2}} \right)}^{2022}}}} = {1 \over {\sqrt 3 }} \times {(2)^{2022}}$

From option (A),

${3^{1010}}\,.\,I\left( {{\pi \over 3}} \right) - I\left( {{\pi \over 6}} \right)$

$= {3^{1010}}\,.\,\sqrt 3 \,.\,{\left( {{2 \over {\sqrt 3 }}} \right)^{2022}} - {{{{(2)}^{2022}}} \over {\sqrt 3 }}$

$= {3^{1010}}\,.\,\sqrt 3 \times {{{2^{2022}}} \over {{3^{1011}}}} - {{{2^{2022}}} \over {\sqrt 3 }}$

$= {{{2^{2022}}} \over {\sqrt 3 }} - {{{2^{2022}}} \over {\sqrt 3 }} = 0$