if `int(1-5sin^2x)/(cos^5xsin^2x)dx=f(x)/(cos^5x)+c` then `f(x)`

Options

A. `-cotx`
B. `-"cosec "x`
C. `"cosec "x`
D. cot x

Explanation

To solve the integral

\[ \int \frac{1 - 5\sin^2 x}{\cos^5 x \sin^2 x} \, dx = \frac{f(x)}{\cos^5 x} + c, \]

we need to find the function \( f(x) \).

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ \int \frac{1 - 5\sin^2 x}{\cos^5 x \sin^2 x} \, dx = \int \left( \frac{1}{\cos^5 x \sin^2 x} - \frac{5\sin^2 x}{\cos^5 x \sin^2 x} \right) \, dx. \]

This simplifies to:

\[ \int \left( \frac{1}{\cos^5 x \sin^2 x} - \frac{5}{\cos^5 x} \right) \, dx. \]

### Step 2: Split the Integral
Now we can split the integral into two parts:

\[ \int \frac{1}{\cos^5 x \sin^2 x} \, dx - 5 \int \frac{1}{\cos^5 x} \, dx. \]

### Step 3: Use Trigonometric Identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( 1 \) as \( \cos^2 x + \sin^2 x \):

\[ \int \left( \frac{\cos^2 x}{\cos^5 x \sin^2 x} + \frac{\sin^2 x}{\cos^5 x \sin^2 x} \right) \, dx = \int \left( \frac{\cos^2 x}{\cos^5 x \sin^2 x} + \frac{1}{\cos^5 x} \right) \, dx. \]

This gives us:

\[ \int \left( \frac{1}{\cos^3 x \sin^2 x} + \frac{1}{\cos^5 x} \right) \, dx. \]

### Step 4: Substitute Variables
Now we will use the substitution \( t = \cos^4 x \sin x \). Then, we differentiate:

\[ dt = (4\cos^3 x (-\sin x) \sin x + \cos^4 x \cos x) \, dx = (-4\cos^3 x \sin^2 x + \cos^4 x \cos x) \, dx. \]

### Step 5: Rewrite the Integral
Rearranging gives us:

\[ \int \frac{dt}{t^2}. \]

### Step 6: Integrate
Using the formula for integration:

\[ \int t^{-2} \, dt = -\frac{1}{t} + c. \]

### Step 7: Back Substitute
Substituting back for \( t \):

\[ -\frac{1}{\cos^4 x \sin x} + c. \]

### Step 8: Final Form
To express this in the required form, we multiply and divide by \( \cos x \):

\[ -\frac{\cos x}{\cos^5 x \sin x} + c = -\cot x \cdot \frac{1}{\cos^5 x} + c. \]

Thus, we find:

\[ f(x) = -\cot x. \]

### Conclusion
The final answer is:

\[ f(x) = -\cot x. \]