The integral `int (dx)/(a cos x + b sin x)` is of the form `(1)/(r) " In" [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))`

Explanation

To solve the integral \(\int \frac{dx}{a \cos x + b \sin x}\) and find the value of \(\alpha\) such that the integral is of the form \(\frac{1}{r} \ln \left[\tan \left(\frac{x + \alpha}{2}\right)\right]\), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{dx}{a \cos x + b \sin x} \]

### Step 2: Express \(a\) and \(b\) in terms of \(r\) and \(\alpha\)
Assume:
\[ a = r \sin \alpha \quad \text{and} \quad b = r \cos \alpha \]
where \(r = \sqrt{a^2 + b^2}\).

### Step 3: Substitute \(a\) and \(b\) into the integral
Substituting \(a\) and \(b\) into the integral gives:
\[ I = \int \frac{dx}{r \sin \alpha \cos x + r \cos \alpha \sin x} \]
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, we can rewrite the denominator:
\[ \sin \alpha \cos x + \cos \alpha \sin x = \sin(\alpha + x) \]
Thus, we have:
\[ I = \frac{1}{r} \int \frac{dx}{\sin(\alpha + x)} \]

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

### Step 6: Identify \(\alpha\)
From the form of the integral, we can see that it matches:
\[ \frac{1}{r} \ln \left[\tan\left(\frac{x + \alpha}{2}\right)\right] \]
Thus, we can conclude that:
\[ \alpha = \tan^{-1}\left(\frac{a}{b}\right) \]

### Final Answer
Therefore, the value of \(\alpha\) is:
\[ \alpha = \tan^{-1}\left(\frac{a}{b}\right) \]