To solve the integral
\[
\int \frac{\sec^2 x - 2010}{\sin^{2010} x} \, dx = \frac{P(x)}{\sin^{2010} x} + C,
\]
we need to find the function \( P(x) \) and then evaluate \( P\left(\frac{\pi}{3}\right) \).
### Step 1: Separate the integral
We can separate the integral into two parts:
\[
\int \frac{\sec^2 x}{\sin^{2010} x} \, dx - 2010 \int \frac{1}{\sin^{2010} x} \, dx.
\]
Let’s denote these two integrals as \( I_1 \) and \( I_2 \):
\[
I_1 = \int \frac{\sec^2 x}{\sin^{2010} x} \, dx,
\]
\[
I_2 = \int \frac{1}{\sin^{2010} x} \, dx.
\]
### Step 2: Solve \( I_1 \)
For \( I_1 \), we can use integration by parts. Let:
- \( u = \sin^{-2010} x \) (which implies \( du = -2010 \sin^{-2011} x \cos x \, dx \))
- \( dv = \sec^2 x \, dx \) (which implies \( v = \tan x \))
Using integration by parts:
\[
I_1 = u v - \int v \, du,
\]
we get:
\[
I_1 = \sin^{-2010} x \tan x - \int \tan x \left(-2010 \sin^{-2011} x \cos x\right) \, dx.
\]
This simplifies to:
\[
I_1 = \frac{\tan x}{\sin^{2010} x} + 2010 \int \frac{\tan x \cos x}{\sin^{2011} x} \, dx.
\]
### Step 3: Solve \( I_2 \)
The integral \( I_2 \) can be written as:
\[
I_2 = \int \csc^{2010} x \, dx.
\]
This integral can be complex, but we can keep it in this form for now.
### Step 4: Combine results
Now we combine the results of \( I_1 \) and \( I_2 \):
\[
\int \frac{\sec^2 x - 2010}{\sin^{2010} x} \, dx = \left(\frac{\tan x}{\sin^{2010} x} + 2010 \int \frac{\tan x \cos x}{\sin^{2011} x} \, dx\right) - 2010 \int \csc^{2010} x \, dx.
\]
### Step 5: Identify \( P(x) \)
From the original equation, we can see that:
\[
P(x) = \tan x.
\]
### Step 6: Evaluate \( P\left(\frac{\pi}{3}\right) \)
Now we need to evaluate \( P\left(\frac{\pi}{3}\right) \):
\[
P\left(\frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.
\]
### Final Answer
Thus, the value of \( P\left(\frac{\pi}{3}\right) \) is
\[
\sqrt{3}.
\]
