The integral `int (dx)/(acosx+bsinx)` is of the form `(1)/(r )ln[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))`

Answer

To solve the integral \(\int \frac{dx}{a \cos x + b \sin x}\) and find the value of \(r\) 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: Use a Trigonometric Identity
We can express \(a\) and \(b\) in terms of \(r\) and an angle \(\alpha\):
Let \(a = r \sin \alpha\) and \(b = r \cos \alpha\). This leads to:
\[ I = \int \frac{dx}{r \sin \alpha \cos x + r \cos \alpha \sin x} \]
This simplifies to:
\[ I = \int \frac{dx}{r(\sin \alpha \cos x + \cos \alpha \sin x)} \]

### Step 3: Simplify the Denominator
Using the sine addition formula, 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 4: Integrate
The integral of \(\frac{1}{\sin(x + \alpha)}\) is:
\[ \int \frac{dx}{\sin(x + \alpha)} = \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| + C \]
Therefore, we have:
\[ I = \frac{1}{r} \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| + C \]

### Step 5: Identify \(r\)
From the form of the integral, we can see that:
\[ \frac{1}{r} \ln \left| \tan \left( \frac{x + \alpha}{2} \right) \right| \]
This implies that \(r\) is equal to:
\[ r = \sqrt{a^2 + b^2} \]

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