I was reading "Inside Interesting Integrals" by Paul J. Nahin. In Section 2.3, this integral has been solved. $$I_n(\alpha)=\int_0^\pi{\dfrac{\cos(n\theta)-\cos(n\alpha)}{\cos\theta-\cos\alpha}}d\theta$$ Using some manipulations, this recursive equation was obtained. $$I_{n+1}(\alpha)-2\cos\alpha I_n(\alpha)+I_{n-1}(\alpha)=0$$ The author has directly used the solutions of this equation.
It is well-known that such a so-called recursive equation has solutions of the form $I_n=Ce^{sn}$ where $C$ and $s$ are constants.
I want to know how this solution is derived.
Also, we obtain $s=\pm i\alpha$ by putting the solution in the equation. Using this, it is stated that $I_n(\alpha)=C_1e^{in\alpha}+C_2e^{-in\alpha}$. But, shouldn't it be just $I_n(\alpha)=Ce^{\pm in\alpha}$?
The given is a recurrence relation, of the form $a_n = Aa_{n-1} + Ba_{n-2}$ which is an example of a recurrence relation of second order with constant coefficients . In your case $A = 2\cos\alpha$ and $B = -1$. Essentially the solutions to recurrence relations and differential equations first follow by using educated guesses.
As for why we add the two, the solution we represent is the general solution, i.e. since the two are indeed solutions for the recurrence relation, any linear combination
$$ C_1 e^{ina} + C_2 e^{-ina}$$
of the two will be a valid solution for our relation. However, if we are given initial conditions, then we can just use the conditions to either find the relation between $C1$ and $C2$ (if 1 conditions) or determine the coefficients themselves (if 2 conditions given).