The integral `int (dx)/(acosx+bsinx)` is of the form `(1)/(r )ln[tan((x+alpha)/(2))]`
What is `alpha` equal to ?

Options

A. `tan^(-1)((a)/(b))`
B. `tan^(-1)((b)/(a))`
C. `tan^(-1)((a+b)/(a-b))`
D. `tan^(-1)((a-b)/(a+b))`

Answer

To solve the integral \(\int \frac{dx}{A \cos x + B \sin x}\) and find the value of \(\alpha\) in the expression \(\frac{1}{R} \ln \left( \tan \left( \frac{x + \alpha}{2} \right) \right)\), we can follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int \frac{dx}{A \cos x + B \sin x} \]

### Step 2: Substitute \(A\) and \(B\)
We can express \(A\) and \(B\) in terms of \(R\) and \(\alpha\):
\[ A = R \sin \alpha \quad \text{and} \quad B = R \cos \alpha \]
This gives us:
\[ I = \int \frac{dx}{R \sin \alpha \cos x + R \cos \alpha \sin x} \]

### Step 3: Factor out \(R\)
Factoring out \(R\) from the denominator:
\[ I = \frac{1}{R} \int \frac{dx}{\sin \alpha \cos x + \cos \alpha \sin x} \]

### Step 4: Use the sine addition formula
Using the sine addition formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), we can rewrite the denominator:
\[ \sin \alpha \cos x + \cos \alpha \sin x = \sin(x + \alpha) \]
Thus, we have:
\[ I = \frac{1}{R} \int \frac{dx}{\sin(x + \alpha)} \]

### Step 5: Integrate
The integral of \(\frac{1}{\sin(x + \alpha)}\) is:
\[ \int \frac{dx}{\sin(x + \alpha)} = \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| + C \]
So, we can write:
\[ I = \frac{1}{R} \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| + C \]

### Step 6: Find \(\alpha\)
From our substitutions, we know:
\[ \tan \alpha = \frac{A}{B} \]
Thus, we can express \(\alpha\) as:
\[ \alpha = \tan^{-1} \left( \frac{A}{B} \right) \]

### Conclusion
The value of \(\alpha\) is:
\[ \alpha = \tan^{-1} \left( \frac{A}{B} \right) \]