1. **State the problem:** Evaluate the integral $$I=\int_0^{\frac{\pi}{2}} \sin^5 \theta \cos^4 \theta \, d\theta.$$\n\n2. **Formula and approach:** When integrating powers of sine and cosine, if one power is odd, we can use substitution by separating one sine factor and converting the rest using $\sin^2 \theta = 1 - \cos^2 \theta$. Here, $\sin^5 \theta = \sin^4 \theta \sin \theta = (\sin^2 \theta)^2 \sin \theta$.\n\n3. **Rewrite the integral:**\n$$I = \int_0^{\frac{\pi}{2}} (\sin^2 \theta)^2 \cos^4 \theta \sin \theta \, d\theta = \int_0^{\frac{\pi}{2}} (1 - \cos^2 \theta)^2 \cos^4 \theta \sin \theta \, d\theta.$$\n\n4. **Substitution:** Let $u = \cos \theta$, then $du = -\sin \theta \, d\theta$ or $-du = \sin \theta \, d\theta$. The limits change as $\theta=0 \Rightarrow u=1$ and $\theta=\frac{\pi}{2} \Rightarrow u=0$.\n\n5. **Rewrite integral in terms of $u$:**\n$$I = \int_{u=1}^{0} (1 - u^2)^2 u^4 (-du) = \int_0^1 (1 - u^2)^2 u^4 \, du.$$\n\n6. **Expand the integrand:**\n$$(1 - u^2)^2 u^4 = (1 - 2u^2 + u^4) u^4 = u^4 - 2u^6 + u^8.$$\n\n7. **Integral becomes:**\n$$I = \int_0^1 (u^4 - 2u^6 + u^8) \, du = \int_0^1 u^4 \, du - 2 \int_0^1 u^6 \, du + \int_0^1 u^8 \, du.$$\n\n8. **Evaluate each integral:**\n$$\int_0^1 u^n \, du = \frac{1}{n+1}.$$\nSo,\n$$\int_0^1 u^4 \, du = \frac{1}{5}, \quad \int_0^1 u^6 \, du = \frac{1}{7}, \quad \int_0^1 u^8 \, du = \frac{1}{9}.$$\n\n9. **Combine results:**\n$$I = \frac{1}{5} - 2 \times \frac{1}{7} + \frac{1}{9} = \frac{1}{5} - \frac{2}{7} + \frac{1}{9}.$$\n\n10. **Find common denominator and simplify:**\nThe common denominator is 315.\n$$\frac{1}{5} = \frac{63}{315}, \quad \frac{2}{7} = \frac{90}{315}, \quad \frac{1}{9} = \frac{35}{315}.$$\nSo,\n$$I = \frac{63}{315} - \frac{90}{315} + \frac{35}{315} = \frac{63 - 90 + 35}{315} = \frac{8}{315}.$$\n\n**Final answer:**\n$$\boxed{\frac{8}{315}}.$$