`int(cosx)/((1+sinx)(2+sin x))dx`

Explanation

To solve the integral \( \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = 1 + \sin x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = \cos x \, dx \]
This means that \( \cos x \, dx = dt \).

### Step 2: Rewrite the Integral
Now, we can rewrite the integral in terms of \( t \):
\[ \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \, dx = \int \frac{1}{t(1 + t)} \, dt \]
Here, we have replaced \( 1 + \sin x \) with \( t \) and \( 2 + \sin x \) with \( 1 + t \).

### Step 3: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on \( \frac{1}{t(1 + t)} \):
\[ \frac{1}{t(1 + t)} = \frac{A}{t} + \frac{B}{1 + t} \]
Multiplying through by \( t(1 + t) \) gives:
\[ 1 = A(1 + t) + Bt \]
Expanding and rearranging, we have:
\[ 1 = A + At + Bt \]
Combining like terms:
\[ 1 = A + (A + B)t \]
From this, we can set up the equations:
1. \( A = 1 \)
2. \( A + B = 0 \)

From the first equation, \( A = 1 \). Substituting into the second equation gives:
\[ 1 + B = 0 \implies B = -1 \]
Thus, we can write:
\[ \frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t} \]

### Step 4: Integrate
Now we can integrate:
\[ \int \left( \frac{1}{t} - \frac{1}{1 + t} \right) dt = \int \frac{1}{t} \, dt - \int \frac{1}{1 + t} \, dt \]
Calculating these integrals gives:
\[ \ln |t| - \ln |1 + t| + C \]

### Step 5: Substitute Back
Now we substitute back \( t = 1 + \sin x \):
\[ \ln |1 + \sin x| - \ln |2 + \sin x| + C \]
Using the property of logarithms \( \ln a - \ln b = \ln \frac{a}{b} \):
\[ \ln \left( \frac{1 + \sin x}{2 + \sin x} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \, dx = \ln \left( \frac{1 + \sin x}{2 + \sin x} \right) + C \]