If `int(sinx)/(cosx(1+cosx))dx=f(x)+C`, then f(x) is equal to

Options

A. `log|(1+cosx)/(cosx)|`
B. `log|(cosx)/(1+cosx)|`
C. `log|(sinx)/(1+sinx)|`
D. `log|(1+sinx)/(sinx)|`

Explanation

To solve the integral \( \int \frac{\sin x}{\cos x (1 + \cos x)} \, dx \) and find \( f(x) \) such that \( \int \frac{\sin x}{\cos x (1 + \cos x)} \, dx = f(x) + C \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\cos x (1 + \cos x)} \, dx \]

### Step 2: Substitution
We use the substitution \( t = \cos x \). Then, the derivative \( dt = -\sin x \, dx \) implies \( \sin x \, dx = -dt \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{-dt}{t(1 + t)} \]

### Step 3: Simplifying the Integral
This can be rewritten as:
\[ I = -\int \frac{dt}{t(1 + t)} \]
Next, we can use partial fraction decomposition:
\[ \frac{1}{t(1 + t)} = \frac{A}{t} + \frac{B}{1 + t} \]
Multiplying through by \( t(1 + t) \) gives:
\[ 1 = A(1 + t) + Bt \]
Expanding and rearranging:
\[ 1 = A + At + Bt \implies 1 = A + (A + B)t \]
From this, we can set up the equations:
1. \( A = 1 \)
2. \( A + B = 0 \)

Solving these gives \( A = 1 \) and \( B = -1 \). Thus, we can write:
\[ \frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t} \]

### Step 4: Integrate Each Term
Now, we can integrate:
\[ I = -\int \left( \frac{1}{t} - \frac{1}{1 + t} \right) dt = -\left( \ln |t| - \ln |1 + t| \right) + C \]
This simplifies to:
\[ I = -\ln \left| \frac{t}{1 + t} \right| + C \]

### Step 5: Substitute Back
Substituting back \( t = \cos x \):
\[ I = -\ln \left| \frac{\cos x}{1 + \cos x} \right| + C \]
This can be rewritten using properties of logarithms:
\[ I = \ln \left| \frac{1 + \cos x}{\cos x} \right| + C \]

### Step 6: Final Result
Thus, we find:
\[ f(x) = \ln \left( \frac{1 + \cos x}{\cos x} \right) \]