If `int(1)/((sinx+4)(sinx-1))dx`
`=A(1)/("tan"(x)/(2)-1)+B"tan"^(-1){f(x)}+C`. Then,

Options

A. `A = (1)/(5), B = (-2)/(5sqrt(15)), f(x) = (4 tan x + 3)/(sqrt(15))`
B. `A = - (1)/(5), B tan^(-1), B = (1)/(sqrt(15)), f(x) = ( 4 tan ((x)/(2)) + 1)/(sqrt(15))`
C. `A = (2)/(5), B = (-2)/(5),f(x) = (4 tan x + 1)/(5)`
D. `A = (2)/(5),B = (-2)/(5sqrt(15)), f(x) = (4 tan (x)/(2) + 1)/(sqrt(15))`

Answer

To solve the integral \( \int \frac{1}{(\sin x + 4)(\sin x - 1)} \, dx \), we will use the method of partial fractions.

### Step 1: Set up the partial fractions
We can express the integrand as:
\[ \frac{1}{(\sin x + 4)(\sin x - 1)} = \frac{A}{\sin x + 4} + \frac{B}{\sin x - 1} \]
where \( A \) and \( B \) are constants to be determined.

### Step 2: Multiply through by the denominator
Multiplying both sides by \( (\sin x + 4)(\sin x - 1) \) gives:
\[ 1 = A(\sin x - 1) + B(\sin x + 4) \]

### Step 3: Expand and collect terms
Expanding the right-hand side:
\[ 1 = A \sin x - A + B \sin x + 4B \]
Combining like terms:
\[ 1 = (A + B) \sin x + (4B - A) \]

### Step 4: Set up the system of equations
For the equation to hold for all \( x \), the coefficients of \( \sin x \) must match and the constant terms must match:
1. \( A + B = 0 \)
2. \( 4B - A = 1 \)

### Step 5: Solve the system of equations
From the first equation, we can express \( A \) in terms of \( B \):
\[ A = -B \]
Substituting into the second equation:
\[ 4B - (-B) = 1 \implies 4B + B = 1 \implies 5B = 1 \implies B = \frac{1}{5} \]
Now substituting back to find \( A \):
\[ A = -B = -\frac{1}{5} \]

### Step 6: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{1}{(\sin x + 4)(\sin x - 1)} \, dx = \int \left( \frac{-\frac{1}{5}}{\sin x + 4} + \frac{\frac{1}{5}}{\sin x - 1} \right) \, dx \]
This simplifies to:
\[ -\frac{1}{5} \int \frac{1}{\sin x + 4} \, dx + \frac{1}{5} \int \frac{1}{\sin x - 1} \, dx \]

### Step 7: Integrate each term
The integrals can be solved using standard techniques. We will denote:
\[ I_1 = \int \frac{1}{\sin x + 4} \, dx \quad \text{and} \quad I_2 = \int \frac{1}{\sin x - 1} \, dx \]
Thus, we have:
\[ \int \frac{1}{(\sin x + 4)(\sin x - 1)} \, dx = -\frac{1}{5} I_1 + \frac{1}{5} I_2 + C \]

### Step 8: Identify constants A, B, and f(x)
From the expression, we can identify:
- \( A = -\frac{1}{5} \)
- \( B = \frac{1}{5} \)
- \( f(x) = \sin x \) (as we are integrating with respect to \( \sin x \))

### Final Answer
Thus, the values are:
- \( A = -\frac{1}{5} \)
- \( B = \frac{1}{5} \)
- \( f(x) = \sin x \)