`int(1-7cos^(2)x)/(sin^(7)xcos^(2)x)dx=(f(x))/((sinx)^(7))+C`, then f(x) is equal to

Options

A. `sinx`
B. `cos x`
C. `tanx`
D. `cotx`

Explanation

To solve the integral

\[ I = \int \frac{1 - 7 \cos^2 x}{\sin^7 x \cos^2 x} \, dx, \]

we can start by breaking it down into two separate integrals:

\[ I = \int \frac{1}{\sin^7 x \cos^2 x} \, dx - 7 \int \frac{\cos^2 x}{\sin^7 x \cos^2 x} \, dx. \]

This simplifies to:

\[ I = \int \frac{1}{\sin^7 x \cos^2 x} \, dx - 7 \int \frac{1}{\sin^7 x} \, dx. \]

Let's denote the first integral as \( I_1 \) and the second integral as \( I_2 \):

\[ I_1 = \int \frac{1}{\sin^7 x \cos^2 x} \, dx, \]
\[ I_2 = \int \frac{1}{\sin^7 x} \, dx. \]

### Step 1: Solve \( I_1 \)

For \( I_1 \), we can use integration by parts. Let:

- \( u = \frac{1}{\sin^7 x} \)
- \( dv = \sec^2 x \, dx \) (since \( \frac{1}{\cos^2 x} = \sec^2 x \))

Then we differentiate and integrate:

- \( du = -7 \frac{\cos x}{\sin^8 x} \, dx \)
- \( v = \tan x \)

Using integration by parts:

\[ I_1 = uv - \int v \, du, \]
\[ I_1 = \frac{1}{\sin^7 x} \tan x - \int \tan x \left(-7 \frac{\cos x}{\sin^8 x}\right) \, dx. \]

This simplifies to:

\[ I_1 = \frac{\tan x}{\sin^7 x} + 7 \int \frac{\tan x \cos x}{\sin^8 x} \, dx. \]

### Step 2: Solve \( I_2 \)

Now, for \( I_2 \):

\[ I_2 = \int \frac{1}{\sin^7 x} \, dx. \]

This integral can be solved using a known formula or further substitution, but for our purposes, we will keep it as is for now.

### Step 3: Combine Results

Now we can combine the results:

\[ I = I_1 - 7 I_2. \]

Substituting back:

\[ I = \left(\frac{\tan x}{\sin^7 x} + 7 \int \frac{\tan x \cos x}{\sin^8 x} \, dx\right) - 7 \int \frac{1}{\sin^7 x} \, dx. \]

### Step 4: Final Expression

We are given that:

\[ I = \frac{f(x)}{\sin^7 x} + C. \]

From our expression, we can see that \( f(x) \) must equal the terms we have derived. Thus, we can conclude:

\[ f(x) = \tan x. \]

### Final Answer

Therefore, the value of \( f(x) \) is:

\[ \boxed{\tan x}. \]