The integral `int (dx)/(a cos x + b sin x)` is of the form `(1)/(r) " In" [tan ((x + alpha)/(2))]`
What is r equal to ?

Options

A. `a^(2) + b^(2)`
B. `sqrt(a^(2) + b^(2))`
C. `a + b`
D. `sqrt(a^(2) + b^(2))`

Explanation

To solve the integral \( \int \frac{dx}{a \cos x + b \sin x} \) and find the value of \( r \) in the expression \( \frac{1}{r} \ln \left( \tan \left( \frac{x + \alpha}{2} \right) \right) \), 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: Use Trigonometric Identity
We can express \( a \cos x + b \sin x \) in a different form using the angle addition formula. We can write:
\[ a \cos x + b \sin x = r \sin(\alpha) \cos x + r \cos(\alpha) \sin x \]
where \( r = \sqrt{a^2 + b^2} \) and \( \sin(\alpha) = \frac{b}{r} \), \( \cos(\alpha) = \frac{a}{r} \).

### Step 3: Substitute in the Integral
Now we substitute this into the integral:
\[ I = \int \frac{dx}{r \sin(\alpha) \cos x + r \cos(\alpha) \sin x} \]
This simplifies to:
\[ I = \frac{1}{r} \int \frac{dx}{\sin(\alpha) \cos x + \cos(\alpha) \sin x} \]

### Step 4: Simplify the Integral
Using the identity \( \sin(A + B) = \sin A \cos B + \cos A \sin B \), we can rewrite the denominator:
\[ \sin(\alpha) \cos x + \cos(\alpha) \sin x = \sin(x + \alpha) \]
Thus, the integral becomes:
\[ I = \frac{1}{r} \int \frac{dx}{\sin(x + \alpha)} \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{dx}{\sin(x + \alpha)} \) can be evaluated using the logarithmic identity:
\[ \int \frac{dx}{\sin x} = \ln \left| \tan \left( \frac{x}{2} + \frac{\alpha}{2} \right) \right| + C \]
So we have:
\[ I = \frac{1}{r} \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| + C \]

### Step 6: Conclusion
From the form of the integral, we can conclude that:
\[ r = \sqrt{a^2 + b^2} \]

Thus, the value of \( r \) is:
\[ \boxed{\sqrt{a^2 + b^2}} \]