For [I(x)= ^2 x-2022 ^2022 x d x], if [I(()/(4))=2^1011], then

Question

For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then

Accepted Answer

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),

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

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

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

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

Solutions · 1