1. **Problem Statement:** Find a reduction formula for the integral $$\int \sin^n x \, dx$$ where $n$ is a positive integer. 2. **Formula and Rules:** We use integration by parts and the identity $$\sin^2 x = 1 - \cos^2 x$$ to reduce the power $n$ step-by-step. 3. **Step 1: Express the integral** $$I_n = \int \sin^n x \, dx$$ 4. **Step 2: Use integration by parts** Let $$u = \sin^{n-1} x$$ and $$dv = \sin x \, dx$$. Then $$du = (n-1) \sin^{n-2} x \cos x \, dx$$ and $$v = -\cos x$$. 5. **Step 3: Apply integration by parts formula** $$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx$$ 6. **Step 4: Use identity for $$\cos^2 x$$** $$\cos^2 x = 1 - \sin^2 x$$, so $$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx$$ 7. **Step 5: Split the integral** $$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \, dx - (n-1) \int \sin^n x \, dx$$ 8. **Step 6: Rearrange to isolate $$I_n$$** $$I_n + (n-1) I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$$ 9. **Step 7: Simplify** $$n I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$$ 10. **Final reduction formula:** $$\boxed{I_n = \frac{-\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2}}$$ This formula expresses the integral of $$\sin^n x$$ in terms of the integral of $$\sin^{n-2} x$$, reducing the power by 2 each time. This is useful for evaluating integrals with high powers of sine by repeated application.