`int(cosx)/((1-sinx)^(3)(2+sinx))dx=`

Options

A. `-(1)/(27)log |1-sinx|-(1)/(9(1-sinx))+(1)/(6(1-sinx)^(2))-(1)/(27)log|2+sinx|+c`
B. `(1)/(27)log|1+sinx|-(1)/(9(1+sinx))-(1)/(6(1-sinx)^(2))+(1)/(27)log|2+sinx|+c`
C. `-(1)/(27)log|1+sinx|-(1)/(9(1-sinx))-(1)/(6(1-sinx)^(2))+(1)/(27)log|2+sinx|+c`
D. `-(1)/(27)log|1-sinx|+(1)/(9(1-sinx))+(1)/(6(1-sinx)^(2))+(1)/(27)log|2+sinx|+c`

Answer

To solve the integral

\[ \int \frac{\cos x}{(1 - \sin x)^3 (2 + \sin x)} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( \sin x = t \). Then, we differentiate to find \( \cos x \, dx = dt \). Thus, we can rewrite the integral as:

\[ \int \frac{dt}{(1 - t)^3 (2 + t)}. \]

### Step 2: Partial Fraction Decomposition
We need to express the integrand using partial fractions:

\[ \frac{1}{(1 - t)^3 (2 + t)} = \frac{A}{1 - t} + \frac{B}{(1 - t)^2} + \frac{C}{(1 - t)^3} + \frac{D}{2 + t}. \]

Multiplying through by the denominator \( (1 - t)^3 (2 + t) \) gives:

\[ 1 = A(1 - t)^2(2 + t) + B(1 - t)(2 + t) + C(2 + t) + D(1 - t)^3. \]

### Step 3: Finding Coefficients
To find the coefficients \( A, B, C, D \), we can substitute convenient values for \( t \):

1. **Let \( t = 1 \)**:
\[ 1 = D(1 - 1)^3 \Rightarrow D = 0. \]

2. **Let \( t = -2 \)**:
\[ 1 = C(2 - 2) \Rightarrow C = \frac{1}{3}. \]

3. **Let \( t = 0 \)**:
\[ 1 = A(1)(2) + B(1)(2) + C(2) \Rightarrow 2A + 2B + \frac{2}{3} = 1. \]

4. **Let \( t = -1 \)**:
\[ 1 = A(4) + B(0) + C(1) + D(8) \Rightarrow 4A + \frac{1}{3} = 1. \]

Solving these equations will yield the values of \( A, B, C \).

### Step 4: Integrating Each Term
After finding the coefficients, we can integrate each term separately:

\[ \int \left( \frac{A}{1 - t} + \frac{B}{(1 - t)^2} + \frac{C}{(1 - t)^3} + \frac{D}{2 + t} \right) dt. \]

### Step 5: Back Substitution
After integrating, we will substitute back \( t = \sin x \) to express the result in terms of \( x \).

### Final Result
The final result will be a combination of logarithmic and polynomial terms based on the integration results.