1. **Problem:** Evaluate the integral $$\int \sin^4 x \cos^3 x \, dx$$ 2. **Formula and rules:** When integrating powers of sine and cosine, if one power is odd, save one sine or cosine factor and convert the rest using the Pythagorean identity $$\sin^2 x = 1 - \cos^2 x$$ or $$\cos^2 x = 1 - \sin^2 x$$. 3. **Step 1:** Since the power of cosine is odd (3), save one cosine factor: $$\int \sin^4 x \cos^3 x \, dx = \int \sin^4 x \cos^2 x \cos x \, dx$$ 4. **Step 2:** Use $$\cos^2 x = 1 - \sin^2 x$$: $$= \int \sin^4 x (1 - \sin^2 x) \cos x \, dx$$ 5. **Step 3:** Substitute $$u = \sin x$$, so $$du = \cos x \, dx$$: $$= \int u^4 (1 - u^2) \, du = \int (u^4 - u^6) \, du$$ 6. **Step 4:** Integrate term by term: $$= \frac{u^5}{5} - \frac{u^7}{7} + C$$ 7. **Step 5:** Substitute back $$u = \sin x$$: $$= \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$$ **Final answer:** $$\int \sin^4 x \cos^3 x \, dx = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C$$