Calculate the integral: integral from 0 to (2 pi) ((a+

Calculate the integral: $\int_0^{2\pi} \frac{(a+\cos x) \, dx}{1+2a\cos x+a^2}$.

We will split the integral and use a standard trigonometric integral result.

Step 1: Split the integral

\[\int_0^{2\pi} \frac{(a+\cos x) \, dx}{1+2a\cos x+a^2} = a\int_0^{2\pi} \frac{dx}{1+2a\cos x+a^2} + \int_0^{2\pi} \frac{\cos x \, dx}{1+2a\cos x+a^2}\]

Step 2: Relate the numerator to the denominator

Notice that $a+\cos x$ can be rewritten using the denominator $D = 1+2a\cos x+a^2$:

\[a + \cos x = \frac{1}{2a}\left[(1 + 2a\cos x + a^2) + (a^2-1)\right] = \frac{D}{2a} + \frac{a^2-1}{2a}\]

Therefore:

\[\frac{a+\cos x}{D} = \frac{1}{2a} + \frac{a^2-1}{2a} \cdot \frac{1}{D}\]

Step 3: Rewrite the integral

\[\int_0^{2\pi} \frac{(a+\cos x) \, dx}{D} = \frac{1}{2a}\int_0^{2\pi} dx + \frac{a^2-1}{2a}\int_0^{2\pi} \frac{dx}{D}\]

\[= \frac{\pi}{a} + \frac{a^2-1}{2a} \int_0^{2\pi} \frac{dx}{1+2a\cos x+a^2}\]

Step 4: Evaluate $\int_0^{2\pi} \frac{dx}{1+2a\cos x+a^2}$

This is a standard integral. For $|a| \neq 1$:

\[\int_0^{2\pi} \frac{dx}{1+2a\cos x+a^2} = \frac{2\pi}{|1-a^2|}\]

i.e. $\frac{2\pi}{1-a^2}$ for $|a|<1$, and $\frac{2\pi}{a^2-1}$ for $|a|>1$.

Step 5: Combine the results

Case 1: $|a| < 1$

\[\int_0^{2\pi} \frac{(a+\cos x) \, dx}{1+2a\cos x+a^2} = \frac{\pi}{a} + \frac{a^2-1}{2a} \cdot \frac{2\pi}{1-a^2} = \frac{\pi}{a} - \frac{\pi}{a} = 0\]

Case 2: $|a| > 1$

\[\int_0^{2\pi} \frac{(a+\cos x) \, dx}{1+2a\cos x+a^2} = \frac{\pi}{a} + \frac{a^2-1}{2a} \cdot \frac{2\pi}{a^2-1} = \frac{\pi}{a} + \frac{\pi}{a} = \frac{2\pi}{a}\]

Final answer:

\[\boxed{\int_0^{2\pi} \frac{(a+\cos x) \, dx}{1+2a\cos x+a^2} = \begin{cases} 0 & \text{if } |a| < 1 \\ \frac{2\pi}{a} & \text{if } |a| > 1 \end{cases}}\]