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

• A. $3^{1010} I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0$
• B. $3^{1010} I \left(\frac{\pi}{6}\right)- I \left(\frac{\pi}{3}\right)=0$
• C. $3^{1011} I \left(\frac{\pi}{3}\right)= I \left(\frac{\pi}{6}\right)-0$
• D. $3^{1011} I \left(\frac{\pi}{6}\right)- I \left(\frac{\pi}{3}\right)=0$

Answer 1

Answer: (A)

$ I ( x )=\underset{II}{\int \sec ^2 x} \cdot \underset{I}{\sin ^{-2022} xdx} -2022 \int \sin ^{-2022} xdx$
$=tan _{ x } \cdot(\sin x )^{-2022}+\int(2022) \tan x \cdot(\sin x )^{-2023} \cos xdx$
$ -2022 \int(\sin x )^{-2022} dx $
$ I ( x )=(\tan x )(\sin x )^{-2022}+ C $
$ \text { At } X =\pi / 4,2^{1011}=\left(\frac{1}{\sqrt{2}}\right)^{-2022}+ C \therefore C =0 $
$\text { Hence } I ( x )=\frac{\tan x }{(\sin x )^{2022}} $
$ I (\pi / 6)=\frac{1}{\sqrt{3}\left(\frac{1}{2}\right)^{2022}}=\frac{2^{2022}}{\sqrt{3}}$
$ I (\pi / 3)=\frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)^{2022}}=\frac{2^{2022}}{(\sqrt{3})^{2021}}=\frac{1}{3^{1010}} I \left(\frac{\pi}{6}\right) $
$3^{1010} I (\pi / 3)= I (\pi / 6)$