For I(x) = ^2x-2022^2022x,dx, if I(4)=2^1011, then:

Question

For $I(x) = \displaystyle\int\frac{\sec^2x-2022}{\sin^{2022}x}\,dx$, if $I\left(\dfrac\pi4\right)=2^{1011}$, then:

Accepted Answer

Write $\dfrac{\sec^2x-2022}{\sin^{2022}x} = \dfrac1{\sin^{2022}x}\sec^2x - \dfrac{2022}{\sin^{2022}x}$. Integrating the first term by parts (with $\dfrac1{\sin^{2022}x}$ as one factor and $\sec^2x\,dx=d(\tan x)$) produces a term that cancels the second, leaving

$$I(x) = \frac{\tan x}{\sin^{2022}x}+C$$

Using $I(\pi/4)=2^{1011}$: $\dfrac{1}{(1/\sqrt2)^{2022}}+C = 2^{1011}+C = 2^{1011} \implies C=0$.

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

$$I\left(\frac\pi3\right) = \frac{\sqrt3}{(\sqrt3/2)^{2022}} = \sqrt3\cdot\frac{2^{2022}}{3^{1011}},\qquad I\left(\frac\pi6\right) = \frac{1/\sqrt3}{(1/2)^{2022}} = \frac{2^{2022}}{\sqrt3}$$

$$3^{1010}I\left(\frac\pi3\right) = 3^{1010}\cdot\sqrt3\cdot\frac{2^{2022}}{3^{1011}} = \frac{2^{2022}}{\sqrt3} = I\left(\frac\pi6\right)$$

confirming option A.

Solutions · 1