Question

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

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