To solve the integral \( \int \frac{\cos^2 x}{\sin^6 x} \, dx \) and express it in the form \( A \cot^5 x + B \cot^3 x + k \), we will follow these steps:
### Step 1: Rewrite the integral
We start with the integral:
\[
I = \int \frac{\cos^2 x}{\sin^6 x} \, dx
\]
We can rewrite \( \sin^6 x \) as \( \sin^2 x \cdot \sin^4 x \):
\[
I = \int \frac{\cos^2 x}{\sin^2 x \cdot \sin^4 x} \, dx
\]
### Step 2: Use trigonometric identities
Using the identity \( \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \) and \( \frac{1}{\sin^4 x} = \csc^4 x \):
\[
I = \int \cot^2 x \cdot \csc^4 x \, dx
\]
### Step 3: Rewrite \( \csc^4 x \)
We know that \( \csc^2 x = 1 + \cot^2 x \), so:
\[
\csc^4 x = (\csc^2 x)^2 = (1 + \cot^2 x)^2 = 1 + 2\cot^2 x + \cot^4 x
\]
Thus, we can rewrite the integral as:
\[
I = \int \cot^2 x (1 + 2\cot^2 x + \cot^4 x) \, dx
\]
### Step 4: Expand the integral
Expanding the integral gives:
\[
I = \int \cot^2 x \, dx + 2 \int \cot^4 x \, dx
\]
### Step 5: Use substitution
Let \( t = \cot x \). Then, \( \frac{dt}{dx} = -\csc^2 x \) or \( dx = -\frac{dt}{\csc^2 x} = -\frac{dt}{1 + t^2} \).
Now, we can express the integrals in terms of \( t \):
1. For \( \int \cot^2 x \, dx \):
\[
\int t^2 \left(-\frac{dt}{1+t^2}\right) = -\int \frac{t^2}{1+t^2} \, dt
\]
This simplifies to:
\[
-\int \left(1 - \frac{1}{1+t^2}\right) dt = -t + \tan^{-1}(t)
\]
2. For \( \int \cot^4 x \, dx \):
\[
\int t^4 \left(-\frac{dt}{1+t^2}\right) = -\int \frac{t^4}{1+t^2} \, dt
\]
This can be simplified to:
\[
-\int (t^2 - 1 + \frac{1}{1+t^2}) dt
\]
### Step 6: Combine results
Combining the results from the integrals gives us:
\[
I = -\frac{t^3}{3} - t + \tan^{-1}(t) + \text{(terms from cot^4 integral)}
\]
### Step 7: Substitute back
Substituting \( t = \cot x \) back into the expression, we get:
\[
I = -\frac{\cot^3 x}{3} - \cot x + \tan^{-1}(\cot x) + \text{(terms from cot^4 integral)}
\]
### Step 8: Identify coefficients
Comparing with the form \( A \cot^5 x + B \cot^3 x + k \), we find:
- \( A = 0 \)
- \( B = -\frac{1}{3} \)
### Step 9: Calculate \( A + B \)
Thus, \( A + B = 0 - \frac{1}{3} = -\frac{1}{3} \).
### Final Answer:
\[
A + B = -\frac{1}{3}
\]
