The value of `int(cos^(3)x)/(sin^(2)x+sinx)dx` is equal to

Options

A. `log_(e)|sinx|+sinx+C`
B. `log_(e)|sinx|-sinx+C`
C. `-log_(e)|sinx|-sinx+C`
D. `-log_(e)|sinx|+sinx+C`

Explanation

To solve the integral \( \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integrand:
\[ \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx = \int \frac{\cos^3 x}{\sin x (\sin x + 1)} \, dx \]

### Step 2: Express \(\cos^3 x\) in terms of \(\sin x\)
Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can express \(\cos^3 x\) as:
\[ \cos^3 x = \cos x \cdot \cos^2 x = \cos x (1 - \sin^2 x) \]
Thus, the integral becomes:
\[ \int \frac{\cos x (1 - \sin^2 x)}{\sin x (\sin x + 1)} \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ \int \frac{\cos x}{\sin x (\sin x + 1)} \, dx - \int \frac{\cos x \sin^2 x}{\sin x (\sin x + 1)} \, dx \]
This simplifies to:
\[ \int \frac{\cos x}{\sin x (\sin x + 1)} \, dx - \int \frac{\cos x \sin x}{\sin x + 1} \, dx \]

### Step 4: Use substitution
Let \( u = \sin x \). Then, \( du = \cos x \, dx \). The integrals become:
\[ \int \frac{1}{u (u + 1)} \, du - \int \frac{u}{u + 1} \, du \]

### Step 5: Simplify the first integral
For the first integral, we can use partial fraction decomposition:
\[ \frac{1}{u(u + 1)} = \frac{A}{u} + \frac{B}{u + 1} \]
Multiplying through by \( u(u + 1) \) and solving for \( A \) and \( B \):
\[ 1 = A(u + 1) + Bu \]
Setting \( u = 0 \) gives \( A = 1 \), and setting \( u = -1 \) gives \( B = -1 \). Thus:
\[ \frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1} \]
Now, we can integrate:
\[ \int \left( \frac{1}{u} - \frac{1}{u + 1} \right) du = \ln |u| - \ln |u + 1| + C_1 \]

### Step 6: Simplify the second integral
For the second integral:
\[ \int \frac{u}{u + 1} \, du = \int \left( 1 - \frac{1}{u + 1} \right) du = u - \ln |u + 1| + C_2 \]

### Step 7: Combine results
Combining both integrals, we have:
\[ \ln |u| - \ln |u + 1| - (u - \ln |u + 1|) + C \]
This simplifies to:
\[ \ln |u| - u + C \]

### Step 8: Substitute back
Substituting \( u = \sin x \):
\[ \ln |\sin x| - \sin x + C \]

### Final Answer
Thus, the value of the integral is:
\[ \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx = \ln |\sin x| - \sin x + C \]