Choose the correct Answer of the Following Questions : `int_0^pi dx/(a + bcosx)`

Options

A. `pi/(ab)`
B. `pi/(a^2 + b^2)`
C. `pi/(a^2-b^2)`
D. 0

Answer

To solve the integral \( I = \int_0^\pi \frac{dx}{a + b \cos x} \), we will follow a series of steps to simplify and evaluate it.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^\pi \frac{dx}{a + b \cos x} \]

### Step 2: Use the Weierstrass Substitution
We can use the substitution \( t = \tan\left(\frac{x}{2}\right) \). This gives us:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad dx = \frac{2}{1 + t^2} dt \]
The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \pi \), \( t = \tan\left(\frac{\pi}{2}\right) = \infty \)

### Step 3: Substitute into the Integral
Substituting these into the integral gives:
\[ I = \int_0^\infty \frac{2}{1 + t^2} \cdot \frac{1}{a + b \frac{1 - t^2}{1 + t^2}} dt \]
This simplifies to:
\[ I = \int_0^\infty \frac{2}{(1 + t^2)(a + b \frac{1 - t^2}{1 + t^2})} dt \]
Simplifying the denominator:
\[ a + b \frac{1 - t^2}{1 + t^2} = \frac{a(1 + t^2) + b(1 - t^2)}{1 + t^2} = \frac{(a + b) + (a - b)t^2}{1 + t^2} \]
Thus, we have:
\[ I = \int_0^\infty \frac{2(1 + t^2)}{(1 + t^2)((a + b) + (a - b)t^2)} dt = \int_0^\infty \frac{2}{(a + b) + (a - b)t^2} dt \]

### Step 4: Evaluate the Integral
This integral can be evaluated using the formula:
\[ \int_0^\infty \frac{dt}{A + Bt^2} = \frac{\pi}{2\sqrt{AB}} \quad \text{for } A > 0, B > 0 \]
Here, \( A = a + b \) and \( B = a - b \). Thus:
\[ I = 2 \cdot \frac{\pi}{2\sqrt{(a + b)(a - b)}} = \frac{\pi}{\sqrt{(a + b)(a - b)}} \]

### Final Result
The final result for the integral is:
\[ I = \frac{\pi}{\sqrt{a^2 - b^2}} \]