Question

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

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

Hint:

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

\(= \int (\sec^2 x \cdot \sin^{-2022} x - 2022 \sin^{-2022} x) \, dx\)

\(= \sin^{-2022} x \tan x + \int 2022 \sin^{-2023} x \cos x \tan x \, dx - \int 2022 \sin^{-2022} x \, dx + C\)

\(I(x) = \sin^{-2022} x \tan x + c\)

\(\therefore I\left(\frac{\pi}{4}\right) = 2^{1011}\)

\(⇒c=2^{1011}−2^{1011}\)
\(⇒c=0\)

\(I(\frac{\pi}{3}) = (\frac{2}{\sqrt{3}} )^{2022}\sqrt{3}\)

\(I\left(\frac{\pi}{6}\right) = 2^{20221}\frac{1}{\sqrt{3}}\)
So, option (A): \(3^{1010} \cdot I\left(\frac{\pi}{3}\right) - I\left(\frac{\pi}{6}\right) = 0\)
 

Answer 1

The Correct Option is A