1. **Stating the problem:** We want to find the integral $$\int \frac{\sin(x+\alpha)}{\sin(x-\alpha)} \, dx$$ where $\alpha$ is a constant. 2. **Formula and approach:** To solve this integral, we use trigonometric identities and algebraic manipulation. Recall the sine addition formula: $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ 3. **Rewrite numerator and denominator:** $$\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha$$ $$\sin(x-\alpha) = \sin x \cos \alpha - \cos x \sin \alpha$$ 4. **Express the integrand:** $$\frac{\sin(x+\alpha)}{\sin(x-\alpha)} = \frac{\sin x \cos \alpha + \cos x \sin \alpha}{\sin x \cos \alpha - \cos x \sin \alpha}$$ 5. **Divide numerator and denominator by $\cos \alpha$ (assuming $\cos \alpha \neq 0$):** $$= \frac{\sin x + \cos x \tan \alpha}{\sin x - \cos x \tan \alpha}$$ 6. **Set $t = \tan \alpha$ for simplicity:** $$= \frac{\sin x + t \cos x}{\sin x - t \cos x}$$ 7. **Rewrite numerator and denominator in terms of tangent:** Divide numerator and denominator by $\cos x$ (assuming $\cos x \neq 0$): $$= \frac{\tan x + t}{\tan x - t}$$ 8. **Substitute $u = \tan x$, so $du = \sec^2 x \, dx = (1 + u^2) \, dx$:** 9. **Rewrite integral:** $$\int \frac{u + t}{u - t} \, dx = \int \frac{u + t}{u - t} \cdot \frac{du}{1 + u^2}$$ 10. **Simplify the integrand:** $$\frac{u + t}{u - t} = 1 + \frac{2t}{u - t}$$ 11. **Split the integral:** $$\int \frac{u + t}{u - t} \cdot \frac{1}{1 + u^2} \, du = \int \frac{1}{1 + u^2} \, du + 2t \int \frac{1}{(u - t)(1 + u^2)} \, du$$ 12. **The first integral is:** $$\int \frac{1}{1 + u^2} \, du = \arctan u = \arctan(\tan x) = x$$ 13. **The second integral requires partial fractions:** Decompose: $$\frac{1}{(u - t)(1 + u^2)} = \frac{A}{u - t} + \frac{B u + C}{1 + u^2}$$ 14. **Solve for A, B, C:** Multiply both sides by $(u - t)(1 + u^2)$: $$1 = A(1 + u^2) + (B u + C)(u - t)$$ 15. **Expand and group terms:** $$1 = A + A u^2 + B u^2 - B t u + C u - C t$$ Group by powers of $u$: $$1 = (A - C t) + u (C - B t) + u^2 (A + B)$$ 16. **Equate coefficients:** - Constant term: $A - C t = 1$ - Coefficient of $u$: $C - B t = 0$ - Coefficient of $u^2$: $A + B = 0$ 17. **Solve system:** From $A + B = 0$, we get $B = -A$. From $C - B t = 0$, substitute $B = -A$: $$C + A t = 0 \Rightarrow C = -A t$$ From $A - C t = 1$, substitute $C = -A t$: $$A - (-A t) t = A + A t^2 = A(1 + t^2) = 1 \Rightarrow A = \frac{1}{1 + t^2}$$ Then: $$B = -\frac{1}{1 + t^2}, \quad C = -\frac{t}{1 + t^2}$$ 18. **Rewrite the integral:** $$2t \int \left( \frac{A}{u - t} + \frac{B u + C}{1 + u^2} \right) du = 2t \left[ A \int \frac{1}{u - t} du + \int \frac{B u + C}{1 + u^2} du \right]$$ 19. **Integrate each term:** - $$\int \frac{1}{u - t} du = \ln|u - t|$$ - $$\int \frac{B u}{1 + u^2} du = \frac{B}{2} \ln(1 + u^2)$$ - $$\int \frac{C}{1 + u^2} du = C \arctan u$$ 20. **Substitute values:** $$2t \left[ \frac{1}{1 + t^2} \ln|u - t| + \left(-\frac{1}{1 + t^2} \cdot \frac{1}{2} \ln(1 + u^2) \right) + \left(-\frac{t}{1 + t^2} \arctan u \right) \right] + C$$ 21. **Simplify:** $$= \frac{2t}{1 + t^2} \ln|u - t| - \frac{t}{1 + t^2} \ln(1 + u^2) - \frac{2 t^2}{1 + t^2} \arctan u + C$$ 22. **Recall $u = \tan x$ and $t = \tan \alpha$:** 23. **Final answer:** $$\int \frac{\sin(x+\alpha)}{\sin(x-\alpha)} dx = x + \frac{2 \tan \alpha}{1 + \tan^2 \alpha} \ln|\tan x - \tan \alpha| - \frac{\tan \alpha}{1 + \tan^2 \alpha} \ln(1 + \tan^2 x) - \frac{2 \tan^2 \alpha}{1 + \tan^2 \alpha} \arctan(\tan x) + C$$ 24. **Use identity:** $$1 + \tan^2 \theta = \sec^2 \theta$$ and $$\arctan(\tan x) = x$$ (for $x$ in principal domain), so the expression can be simplified further if needed.