1. **Stating the problem:** We want to find the integral $$\int \frac{\sin x}{\sin(x+\alpha)} \, dx$$ where $\alpha$ is a constant. 2. **Formula and approach:** To solve this integral, we use the sine addition formula: $$\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha$$ This allows us to rewrite the denominator. 3. **Rewrite the integral:** $$\int \frac{\sin x}{\sin x \cos \alpha + \cos x \sin \alpha} \, dx = \int \frac{\sin x}{\sin x \cos \alpha + \cos x \sin \alpha} \, dx$$ 4. **Divide numerator and denominator by $\sin x$ (assuming $\sin x \neq 0$):** $$= \int \frac{1}{\cos \alpha + \cot x \sin \alpha} \, dx$$ 5. **Substitute $t = \cot x$, so $dt = -\csc^2 x \, dx$ and $dx = -\frac{dt}{1+t^2}$:** Rewrite the integral in terms of $t$: $$= \int \frac{1}{\cos \alpha + t \sin \alpha} \cdot \left(-\frac{dt}{1+t^2}\right) = - \int \frac{dt}{(1+t^2)(\cos \alpha + t \sin \alpha)}$$ 6. **Partial fraction decomposition:** We want to express $$\frac{1}{(1+t^2)(\cos \alpha + t \sin \alpha)} = \frac{A t + B}{1+t^2} + \frac{C}{\cos \alpha + t \sin \alpha}$$ 7. **Solve for constants $A$, $B$, and $C$:** Multiply both sides by $(1+t^2)(\cos \alpha + t \sin \alpha)$: $$1 = (A t + B)(\cos \alpha + t \sin \alpha) + C(1+t^2)$$ Expand: $$1 = A t \cos \alpha + A t^2 \sin \alpha + B \cos \alpha + B t \sin \alpha + C + C t^2$$ Group terms by powers of $t$: $$1 = t^2 (A \sin \alpha + C) + t (A \cos \alpha + B \sin \alpha) + (B \cos \alpha + C)$$ Equate coefficients to zero except constant term: - Coefficient of $t^2$: $A \sin \alpha + C = 0$ - Coefficient of $t$: $A \cos \alpha + B \sin \alpha = 0$ - Constant term: $B \cos \alpha + C = 1$ 8. **Solve the system:** From first: $C = -A \sin \alpha$ From second: $B = -\frac{A \cos \alpha}{\sin \alpha}$ Substitute into third: $$-\frac{A \cos \alpha}{\sin \alpha} \cos \alpha - A \sin \alpha = 1$$ $$-A \frac{\cos^2 \alpha}{\sin \alpha} - A \sin \alpha = 1$$ Multiply both sides by $\sin \alpha$: $$-A \cos^2 \alpha - A \sin^2 \alpha = \sin \alpha$$ $$-A (\cos^2 \alpha + \sin^2 \alpha) = \sin \alpha$$ Since $\cos^2 \alpha + \sin^2 \alpha = 1$: $$-A = \sin \alpha \implies A = -\sin \alpha$$ Then: $$C = -A \sin \alpha = -(-\sin \alpha) \sin \alpha = \sin^2 \alpha$$ $$B = -\frac{A \cos \alpha}{\sin \alpha} = -\frac{-\sin \alpha \cos \alpha}{\sin \alpha} = \cos \alpha$$ 9. **Rewrite the integral:** $$- \int \left( \frac{A t + B}{1+t^2} + \frac{C}{\cos \alpha + t \sin \alpha} \right) dt = - \int \frac{-\sin \alpha t + \cos \alpha}{1+t^2} dt - \int \frac{\sin^2 \alpha}{\cos \alpha + t \sin \alpha} dt$$ 10. **Integrate each term:** - First term: $$- \int \frac{-\sin \alpha t}{1+t^2} dt = \sin \alpha \int \frac{t}{1+t^2} dt$$ Use substitution $u = 1+t^2$, $du = 2t dt$: $$\int \frac{t}{1+t^2} dt = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|1+t^2| + C$$ - Second term: $$- \int \frac{\cos \alpha}{1+t^2} dt = -\cos \alpha \arctan t + C$$ - Third term: $$- \int \frac{\sin^2 \alpha}{\cos \alpha + t \sin \alpha} dt = - \sin^2 \alpha \int \frac{1}{\cos \alpha + t \sin \alpha} dt$$ Use substitution $v = \cos \alpha + t \sin \alpha$, $dv = \sin \alpha dt$, so $$dt = \frac{dv}{\sin \alpha}$$ Integral becomes: $$- \sin^2 \alpha \int \frac{1}{v} \cdot \frac{dv}{\sin \alpha} = - \sin \alpha \int \frac{1}{v} dv = - \sin \alpha \ln|v| + C = - \sin \alpha \ln|\cos \alpha + t \sin \alpha| + C$$ 11. **Combine all results:** $$= \sin \alpha \cdot \frac{1}{2} \ln|1+t^2| - \cos \alpha \arctan t - \sin \alpha \ln|\cos \alpha + t \sin \alpha| + C$$ 12. **Back-substitute $t = \cot x$ and simplify:** Recall $1 + \cot^2 x = \csc^2 x$ and $\arctan(\cot x) = \frac{\pi}{2} - x$: $$= \frac{\sin \alpha}{2} \ln|\csc^2 x| - \cos \alpha \left( \frac{\pi}{2} - x \right) - \sin \alpha \ln|\cos \alpha + \cot x \sin \alpha| + C$$ Simplify $\ln|\csc^2 x| = -2 \ln|\sin x|$: $$= - \sin \alpha \ln|\sin x| - \cos \alpha \left( \frac{\pi}{2} - x \right) - \sin \alpha \ln|\cos \alpha + \cot x \sin \alpha| + C$$ This is the final answer. **Final answer:** $$\int \frac{\sin x}{\sin(x+\alpha)} \, dx = - \sin \alpha \ln|\sin x| - \cos \alpha \left( \frac{\pi}{2} - x \right) - \sin \alpha \ln|\cos \alpha + \cot x \sin \alpha| + C$$