For I(x) = ∫ ^ 2 x - 2022 ^ 2022 x dx , if I((π)/(4)) = 2 ^ 1011 ,...

Given,

$I\left(x\right) = ∫ \dfrac{\sec^{2}x - 2022}{\sin^{2022}x}dx$

On rearranging we get,

$⇒ I\left(x\right) = ∫ \mathop{\mathop{\sec^{2}x}_{⏟}}_{II} \cdot \mathop{\mathop{\sin^{- 2022}x}_{⏟}}_{I}dx - 2022 ∫ \sin^{- 2022}xdx$

Now using integration by parts we get,

$⇒ I\left(x\right) = \tan x . {\left(\sin x\right)}^{- 2022} + ∫ \left(2022\right)\tan x \cdot {\left(\sin x\right)}^{- 2023}cosxdx - 2022 ∫ {\left(\sin x\right)}^{- 2022}dx$

$⇒ I\left(x\right) = \tan x . {\left(\sin x\right)}^{- 2022} + ∫ \left(2022\right)\tan x \cdot {\left(\sin x\right)}^{- 2022}\dfrac{\cos x}{\sin x}dx - 2022 ∫ {\left(\sin x\right)}^{- 2022}dx$

$⇒ I\left(x\right) = \tan x . {\left(\sin x\right)}^{- 2022} + ∫ \left(2022\right){\left(\sin x\right)}^{- 2022}dx - 2022 ∫ {\left(\sin x\right)}^{- 2022}dx$

$⇒ I\left(x\right) = \left(\tan x\right){\left(\sin x\right)}^{- 2022} + C$

Now at $x = \dfrac{\pi}{4}$

$I\left(\dfrac{\pi}{4}\right) = {2}^{1011} ⇒ {2}^{1011} = 1 \times {\left(\dfrac{1}{\sqrt{2}}\right)}^{- 2022} + C ⇒ C = 0$

Hence $I\left(x\right) = \dfrac{\tan x}{{\left(\sin x\right)}^{2022}}$

Now finding the value of $I\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}{\left(\dfrac{1}{2}\right)}^{2022}} = \dfrac{{2}^{2022}}{\sqrt{3}}$

And $I\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{{\left(\dfrac{\sqrt{3}}{2}\right)}^{2022}} = \dfrac{{2}^{2022}}{{\left(\sqrt{3}\right)}^{2021}} = \dfrac{1}{{3}^{1010}}I\left(\dfrac{\pi}{6}\right)$

So, ${3}^{1010}I\left(\dfrac{\pi}{3}\right) = I\left(\dfrac{\pi}{6}\right)$