1. Given integral

We need to evaluate

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

with the condition

$$I\left(\frac\pi4\right)=2^{1011}.$$

We must then compare $I\left(\frac\pi3\right)$ and $I\left(\frac\pi6\right)$.

---

2. Rewrite the integrand

Notice that

$$\frac{d}{dx}(\cot x)=-\csc^2 x.$$

Let

$$t=\cot x.$$

Then

$$dt=-\csc^2 x\,dx=-\frac{1}{\sin^2 x}dx \quad\Rightarrow\quad dx=-\sin^2 x\,dt.$$

Also,

$$\sin^2 x=\frac{1}{1+\cot^2 x}=\frac{1}{1+t^2},$$

so

$$\sin^{2022}x=\left(\frac{1}{1+t^2}\right)^{1011}.$$

Now simplify the numerator:

$$\sec^2 x-2022=(1+\tan^2 x)-2022.$$

But a better route is to express using $\sin x,\cos x$:

$$\frac{\sec^2 x-2022}{\sin^{2022}x} =\frac{1}{\cos^2 x\sin^{2022}x}-\frac{2022}{\sin^{2022}x}.$$

This looks messy directly, so instead observe a pattern with differentiation.

---

3. Recognize the derivative form

Consider

$$\frac{d}{dx}\left(\frac{\cot x}{\sin^{2020}x}\right).$$

Since

$$\frac{\cot x}{\sin^{2020}x}=\cos x\,\sin^{-2021}x,$$

we differentiate:

$$\frac{d}{dx}(\cos x\sin^{-2021}x) =-\sin x\sin^{-2021}x+\cos x(-2021)\sin^{-2022}x\cos x.$$

So

$$= -\sin^{-2020}x-2021\frac{\cos^2 x}{\sin^{2022}x}.$$

Now use $\cos^2 x=1-\sin^2 x$:

$$-\sin^{-2020}x-2021\frac{1-\sin^2 x}{\sin^{2022}x} = -\sin^{-2020}x-2021\sin^{-2022}x+2021\sin^{-2020}x.$$

Hence

$$\frac{d}{dx}\left(\frac{\cot x}{\sin^{2020}x}\right) =2020\sin^{-2020}x-2021\sin^{-2022}x.$$

This is still not exactly the integrand.

Let us instead try differentiating

$$\tan x\,\sin^{-2021}x.$$

That is cumbersome too.

So let us do a clean substitution.

---

4. Use substitution $u=\cot x$ properly

Let

$$u=\cot x.$$

Then

$$du=-\csc^2 x\,dx=-\frac{dx}{\sin^2 x} \quad\Rightarrow\quad dx=-\sin^2 x\,du.$$

Also,

$$\sec^2 x = 1+\tan^2 x = 1+\frac{1}{u^2} = \frac{u^2+1}{u^2}.$$

And since

$$1+u^2=\csc^2 x=\frac1{\sin^2 x},$$

we get

$$\sin^{2022}x=(1+u^2)^{-1011}.$$

Therefore,

$$\frac{\sec^2 x-2022}{\sin^{2022}x}dx =\left(\frac{u^2+1}{u^2}-2022\right)(1+u^2)^{1011}(-\sin^2 x)du.$$

But

$$\sin^2 x=\frac1{1+u^2},$$

so this becomes

$$-\left(\frac{u^2+1}{u^2}-2022\right)(1+u^2)^{1010}du.$$

Now simplify the bracket:

$$\frac{u^2+1}{u^2}-2022=1+\frac1{u^2}-2022=\frac1{u^2}-2021.$$

Thus

$$I=\int -\left(\frac1{u^2}-2021\right)(1+u^2)^{1010}du.$$

This still looks complicated, so let us search for a derivative of a simpler expression.

---

5. Key observation

Consider

$$F(x)=\cot x\,\csc^{2020}x=\frac{\cot x}{\sin^{2020}x}.$$

Differentiate using product rule in standard trig form:

$$\frac{d}{dx}(\cot x\,\csc^{2020}x) =( -\csc^2 x)\csc^{2020}x+\cot x\cdot 2020\csc^{2019}x(-\csc x\cot x).$$

So

$$= -\csc^{2022}x-2020\cot^2 x\csc^{2020}x.$$

Factor $\csc^{2020}x$:

$$= -\csc^{2020}x(\csc^2 x+2020\cot^2 x).$$

Using

$$\csc^2 x=1+\cot^2 x,$$

we get

$$\csc^2 x+2020\cot^2 x = 1+2021\cot^2 x.$$

Hence

$$\frac{d}{dx}(\cot x\,\csc^{2020}x)= -\csc^{2020}x(1+2021\cot^2 x).$$

Now write in terms of $\sin x,\cos x$:

$$-\frac{1}{\sin^{2020}x}\left(1+2021\frac{\cos^2 x}{\sin^2 x}\right) = -\frac{\sin^2 x+2021\cos^2 x}{\sin^{2022}x}.$$

Since

$$\sin^2 x+2021\cos^2 x = 1+2020\cos^2 x,$$

this becomes

$$-\frac{1+2020\cos^2 x}{\sin^{2022}x}.$$

Now compare with the integrand:

$$\frac{\sec^2 x-2022}{\sin^{2022}x}.$$

Multiply numerator of our derivative expression by $\sec^2 x$ relation:

$$\sec^2 x-2022=\frac{1-2022\cos^2 x}{\cos^2 x}.$$

This does not match directly. So we try another candidate.

---

6. Try differentiating $\tan x\,\csc^{2021}x$

Let

$$F(x)=\tan x\,\csc^{2021}x.$$

Then

$$F'(x)=\sec^2 x\,\csc^{2021}x+\tan x\cdot 2021\csc^{2020}x(-\csc x\cot x).$$

Since $\tan x\cot x=1$,

$$F'(x)=\sec^2 x\,\csc^{2021}x-2021\csc^{2022}x.$$

Factor $\csc^{2021}x$:

$$F'(x)=\csc^{2021}x(\sec^2 x-2021\csc x).$$

Not the same.

Instead, consider

$$F(x)=\frac{\tan x}{\sin^{2021}x}=\sec x\csc^{2020}x?$$

Still awkward.

---

7. A simpler direct rewrite

Observe that

$$\sec^2 x=\frac{1}{\cos^2 x}=1+\tan^2 x=1+\frac{\sin^2 x}{\cos^2 x}.$$

A much better trick is to split:

$$\frac{\sec^2 x-2022}{\sin^{2022}x} =\frac{1}{\cos^2 x\sin^{2022}x}-\frac{2022}{\sin^{2022}x}.$$

Now let

$$u=\cot x=\frac{\cos x}{\sin x}.$$

Then

$$1+u^2=\csc^2 x,$$

so

$$\sin^{-2022}x=(1+u^2)^{1011},$$

and

$$\cos^2 x=u^2\sin^2 x=\frac{u^2}{1+u^2}.$$

Hence

$$\frac{1}{\cos^2 x\sin^{2022}x} =\frac{1+u^2}{u^2}(1+u^2)^{1011} =\frac{(1+u^2)^{1012}}{u^2}.$$

Also,

$$dx=-\frac{du}{1+u^2}.$$

Therefore

$$I=\int \left[\frac{(1+u^2)^{1012}}{u^2}-2022(1+u^2)^{1011}\right]\left(-\frac{du}{1+u^2}\right).$$

So

$$I=-\int \left[\frac{(1+u^2)^{1011}}{u^2}-2022(1+u^2)^{1010}\right]du.$$

Now combine terms:

$$\frac{1+u^2}{u^2}-2022=\frac1{u^2}-2021,$$

thus

$$I=-\int (1+u^2)^{1010}\left(\frac1{u^2}-2021\right)du.$$

But

$$\frac{d}{du}\left(\frac{(1+u^2)^{1011}}{u}\right) =\frac{2022u(1+u^2)^{1010}\cdot u-(1+u^2)^{1011}}{u^2}$$ $$=\frac{(1+u^2)^{1010}(2022u^2-(1+u^2))}{u^2}$$ $$=(1+u^2)^{1010}\frac{2021u^2-1}{u^2} =-(1+u^2)^{1010}\left(\frac1{u^2}-2021\right).$$

This is exactly the integrand in $u$.

Hence,

$$I=\frac{(1+u^2)^{1011}}{u}+C.$$

Substitute back $u=\cot x$:

$$I(x)=\frac{(1+\cot^2 x)^{1011}}{\cot x}+C =\frac{\csc^{2022}x}{\cot x}+C.$$

Since

$$\frac{\csc^{2022}x}{\cot x} =\frac{1}{\sin^{2022}x}\cdot\frac{\sin x}{\cos x} =\frac{1}{\sin^{2021}x\cos x},$$

we get

$$I(x)=\frac{1}{\sin^{2021}x\cos x}+C.$$

---

8. Use the condition $I(\pi/4)=2^{1011}$

At $x=\pi/4$,

$$\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}.$$

So

$$I\left(\frac\pi4\right)=\frac{1}{\left(\frac1{\sqrt2}\right)^{2021}\left(\frac1{\sqrt2}\right)}+C =\frac{1}{\left(\frac1{\sqrt2}\right)^{2022}}+C =2^{1011}+C.$$

Given this equals $2^{1011}$, we get

$$C=0.$$

Therefore,

$$I(x)=\frac{1}{\sin^{2021}x\cos x}.$$

---

9. Compute $I\left(\frac\pi3\right)$ and $I\left(\frac\pi6\right)$

At $x=\frac\pi3$

$$\sin\frac\pi3=\frac{\sqrt3}{2},\qquad \cos\frac\pi3=\frac12.$$

Thus

$$I\left(\frac\pi3\right)=\frac{1}{\left(\frac{\sqrt3}{2}\right)^{2021}\cdot\frac12} =\frac{2}{\left(\frac{\sqrt3}{2}\right)^{2021}}.$$

Now

$$\left(\frac{\sqrt3}{2}\right)^{2021}=\frac{3^{1010}(\sqrt3)}{2^{2021}}.$$

So

$$I\left(\frac\pi3\right)=2\cdot \frac{2^{2021}}{3^{1010}\sqrt3} =\frac{2^{2022}}{3^{1010}\sqrt3}.$$

At $x=\frac\pi6$

$$\sin\frac\pi6=\frac12,\qquad \cos\frac\pi6=\frac{\sqrt3}{2}.$$

Thus

$$I\left(\frac\pi6\right)=\frac{1}{\left(\frac12\right)^{2021}\cdot\frac{\sqrt3}{2}} =\frac{2^{2022}}{\sqrt3}.$$

Therefore,

$$I\left(\frac\pi6\right)=3^{1010}I\left(\frac\pi3\right).$$

Equivalently,

$$3^{1010}I\left(\frac\pi3\right)-I\left(\frac\pi6\right)=0.$$

So Option A is correct.

---

10. Check all options

• A: $3^{1010}I\left(\frac\pi3\right)-I\left(\frac\pi6\right)=0$ ✅
• B: $3^{1010}I\left(\frac\pi6\right)-I\left(\frac\pi3\right)=0$ ❌
• C: $3^{1011}I\left(\frac\pi3\right)-I\left(\frac\pi6\right)=0$ ❌
• D: $3^{1011}I\left(\frac\pi6\right)-I\left(\frac\pi3\right)=0$ ❌

Hence the single correct answer is A.

---

11. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.