1. **Stating the problem:** We want to evaluate the integral $$\int \frac{\sin(x-\alpha)}{\sin(x+\alpha)} \, dx.$$\n\n2. **Rewrite the integrand:** Use the sine subtraction and addition formulas: $$\sin(x-\alpha) = \sin x \cos \alpha - \cos x \sin \alpha,$$ $$\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha.$$\n\n3. **Express the integrand as a ratio:** $$\frac{\sin(x-\alpha)}{\sin(x+\alpha)} = \frac{\sin x \cos \alpha - \cos x \sin \alpha}{\sin x \cos \alpha + \cos x \sin \alpha}.$$\n\n4. **Divide numerator and denominator by $\cos x \sin \alpha$:** $$= \frac{\frac{\sin x \cos \alpha}{\cos x \sin \alpha} - 1}{\frac{\sin x \cos \alpha}{\cos x \sin \alpha} + 1} = \frac{\frac{\sin x}{\cos x} \frac{\cos \alpha}{\sin \alpha} - 1}{\frac{\sin x}{\cos x} \frac{\cos \alpha}{\sin \alpha} + 1}.$$\n\n5. **Set** $$t = \tan x, \quad k = \frac{\cos \alpha}{\sin \alpha} = \cot \alpha.$$ Then the integrand becomes $$\frac{k t - 1}{k t + 1}.$$\n\n6. **Rewrite the integral in terms of $t$:** Since $$dt = \sec^2 x \, dx = (1 + t^2) \, dx,$$ we have $$dx = \frac{dt}{1 + t^2}.$$\n\n7. **Substitute into the integral:** $$\int \frac{k t - 1}{k t + 1} \, dx = \int \frac{k t - 1}{k t + 1} \cdot \frac{dt}{1 + t^2}.$$\n\n8. **Simplify the integrand:** $$I = \int \frac{k t - 1}{(k t + 1)(1 + t^2)} \, dt.$$\n\n9. **Use partial fraction decomposition:** Write $$\frac{k t - 1}{(k t + 1)(1 + t^2)} = \frac{A}{k t + 1} + \frac{B t + C}{1 + t^2}.$$\n\n10. **Multiply both sides by $(k t + 1)(1 + t^2)$:** $$k t - 1 = A(1 + t^2) + (B t + C)(k t + 1).$$\n\n11. **Expand right side:** $$A + A t^2 + B k t^2 + B t + C k t + C.$$\n\n12. **Group terms by powers of $t$:**\n- Coefficient of $t^2$: $A + B k$\n- Coefficient of $t$: $B + C k$\n- Constant term: $A + C$\n\n13. **Match coefficients with left side:**\n- For $t^2$: $0 = A + B k$\n- For $t$: $k = B + C k$\n- For constant: $-1 = A + C$\n\n14. **Solve the system:**\nFrom $0 = A + B k$, we get $A = -B k$.\nFrom $-1 = A + C$, substitute $A$: $-1 = -B k + C$ so $C = -1 + B k$.\nFrom $k = B + C k$, substitute $C$: $k = B + k(-1 + B k) = B + k B k - k = B + B k^2 - k$.\nRearranged: $k + k = B + B k^2$ or $2 k = B (1 + k^2)$ so $B = \frac{2 k}{1 + k^2}\.$\n\n15. **Find $A$ and $C$:**\n$A = -B k = -\frac{2 k^2}{1 + k^2}$\n$C = -1 + B k = -1 + \frac{2 k^2}{1 + k^2} = \frac{- (1 + k^2) + 2 k^2}{1 + k^2} = \frac{-1 - k^2 + 2 k^2}{1 + k^2} = \frac{-1 + k^2}{1 + k^2}$\n\n16. **Rewrite the integral:** $$I = \int \frac{A}{k t + 1} dt + \int \frac{B t + C}{1 + t^2} dt = A \int \frac{dt}{k t + 1} + \int \frac{B t}{1 + t^2} dt + C \int \frac{dt}{1 + t^2}.$$\n\n17. **Integrate each term:**\n- $$\int \frac{dt}{k t + 1} = \frac{1}{k} \ln |k t + 1| + C_1,$$\n- $$\int \frac{t}{1 + t^2} dt = \frac{1}{2} \ln(1 + t^2) + C_2,$$\n- $$\int \frac{dt}{1 + t^2} = \arctan t + C_3.$$\n\n18. **Substitute back:**\n$$I = A \cdot \frac{1}{k} \ln |k t + 1| + B \cdot \frac{1}{2} \ln(1 + t^2) + C \arctan t + C_4.$$\n\n19. **Plug in values of $A, B, C$:**\n$$I = -\frac{2 k^2}{1 + k^2} \cdot \frac{1}{k} \ln |k t + 1| + \frac{2 k}{1 + k^2} \cdot \frac{1}{2} \ln(1 + t^2) + \frac{-1 + k^2}{1 + k^2} \arctan t + C_4.$$\n\n20. **Simplify coefficients:**\n$$I = -\frac{2 k}{1 + k^2} \ln |k t + 1| + \frac{k}{1 + k^2} \ln(1 + t^2) + \frac{k^2 - 1}{1 + k^2} \arctan t + C.$$\n\n21. **Recall substitutions:** $t = \tan x$, $k = \cot \alpha$.\n\n**Final answer:**\n$$\int \frac{\sin(x-\alpha)}{\sin(x+\alpha)} \, dx = -\frac{2 \cot \alpha}{1 + \cot^2 \alpha} \ln |\cot \alpha \tan x + 1| + \frac{\cot \alpha}{1 + \cot^2 \alpha} \ln(1 + \tan^2 x) + \frac{\cot^2 \alpha - 1}{1 + \cot^2 \alpha} \arctan(\tan x) + C.$$\n\nUsing the identity $$1 + \cot^2 \alpha = \csc^2 \alpha,$$ and $$\arctan(\tan x) = x$$ for principal values, the expression can be further simplified if desired.