JEE Main 2022 — Indefinite Integration Question with Solution

Given,

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

On rearranging we get,

$\Rightarrow I\left(x\right)=\int \underset{II}{\underbrace{{\sec }^{2}x}}·\underset{I}{\underbrace{{\sin }^{-2022}x}}dx-2022\int {\sin }^{-2022}xdx$

Now using integration by parts we get,

$\Rightarrow I\left(x\right)=\tan x.{\left(\sin x\right)}^{-2022}+\int \left(2022\right)\tan x·{\left(\sin x\right)}^{-2023}\cos xdx-2022\int {\left(\sin x\right)}^{-2022}dx$

$\Rightarrow I\left(x\right)=\tan x.{\left(\sin x\right)}^{-2022}+\int \left(2022\right)\tan x·{\left(\sin x\right)}^{-2022}\frac{\cos x}{\sin x}dx-2022\int {\left(\sin x\right)}^{-2022}dx$

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

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

Now at  $x=\frac{\pi }{4}$

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

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

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

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

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