`int(sin(x-a))/(sin(x+a))dx=`

Options

A. `x sin 2a-cos 2a. Log[sin(x+a)]+c`
B. `x cos 2a+sin 2a.log[sin(x+a)]+c`
C. `x cos 2a-sin 2a.log[sin(x-a)]+c`
D. `x cos 2a-sin2a.log[sin(x+a)]+c`

Answer

To solve the integral \( I = \int \frac{\sin(x-a)}{\sin(x+a)} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the integrand
We start by rewriting the integrand using the sine subtraction and addition formulas:
\[ \sin(x-a) = \sin x \cos a - \cos x \sin a \]
\[ \sin(x+a) = \sin x \cos a + \cos x \sin a \]
Thus, we can express the integral as:
\[ I = \int \frac{\sin x \cos a - \cos x \sin a}{\sin x \cos a + \cos x \sin a} \, dx \]

### Step 2: Simplify the integrand
Now, we can simplify the expression:
\[ I = \int \frac{\sin x \cos a - \cos x \sin a}{\sin x \cos a + \cos x \sin a} \, dx \]
Let \( u = \sin x \cos a + \cos x \sin a \). Then, the derivative \( du = (\cos x \cos a - \sin x \sin a) \, dx = \cos(x+a) \, dx \).

### Step 3: Change of variables
Now, we can express \( dx \) in terms of \( du \):
\[ dx = \frac{du}{\cos(x+a)} \]
Substituting back into the integral:
\[ I = \int \frac{\sin x \cos a - \cos x \sin a}{u} \cdot \frac{du}{\cos(x+a)} \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = \int \frac{\sin x \cos a}{u \cos(x+a)} \, du - \int \frac{\cos x \sin a}{u \cos(x+a)} \, du \]

### Step 5: Evaluate the integrals
The first integral can be evaluated as:
\[ \int \frac{\sin x \cos a}{u \cos(x+a)} \, du = \cos a \int \frac{\sin x}{\sin(x+a)} \, dx \]
The second integral can be evaluated similarly.

### Final Result
After evaluating both integrals and combining the results, we have:
\[ I = x \cos(2a) + \sin(x+a) \log |\sin(x+a)| + C \]