1. The problem is to find the integral of $\sin^3(x) \cos^5(x) \, dx$. 2. First, express the powers in a manageable form: rewrite $\sin^3(x)$ as $\sin(x) \sin^2(x)$. 3. Use the identity $\sin^2(x) = 1 - \cos^2(x)$ to get $\sin(x)(1 - \cos^2(x)) \cos^5(x) \, dx$. 4. Substitute $u = \cos(x)$, so $du = -\sin(x) \, dx$, implying $-du = \sin(x) \, dx$. 5. Rewrite the integral in terms of $u$: $$\int \sin(x)(1 - \cos^2(x)) \cos^5(x) \, dx = \int (1 - u^2) u^5 (-du) = -\int (1 - u^2) u^5 \, du$$ 6. Simplify the integrand: $-\int u^5 - u^7 \, du = -\int u^5 \, du + \int u^7 \, du$ 7. Integrate term by term: $$-\frac{u^6}{6} + \frac{u^8}{8} + C$$ 8. Substitute back $u=\cos(x)$ to get the final answer: $$-\frac{\cos^6(x)}{6} + \frac{\cos^8(x)}{8} + C$$.