Evaluate the integral integral 5 x 4 x \,dx.

Since the power of $\sin x$ is odd, we can separate one $\sin x$ term and express the remaining even power of $\sin x$ in terms of $\cos x$.

$$\int \sin^5 x \cos^4 x \,dx = \int \sin^4 x \cos^4 x \sin x \,dx$$ $$= \int (\sin^2 x)^2 \cos^4 x \sin x \,dx$$ $$= \int (1 - \cos^2 x)^2 \cos^4 x \sin x \,dx$$

Let $u = \cos x$.
Then, the differential $du = -\sin x \,dx$.
This means $\sin x \,dx = -du$.

Substitute $u$ and $du$ into the integral:

$$\int (1 - u^2)^2 u^4 (-du)$$ $$= -\int (1 - 2u^2 + u^4) u^4 \,du$$ $$= -\int (u^4 - 2u^6 + u^8) \,du$$

$$= - \left( \frac{u^{4+1}}{4+1} - 2\frac{u^{6+1}}{6+1} + \frac{u^{8+1}}{8+1} \right) + C$$ $$= - \left( \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} \right) + C$$ $$= -\frac{u^5}{5} + \frac{2u^7}{7} - \frac{u^9}{9} + C$$

$$= -\frac{\cos^5 x}{5} + \frac{2\cos^7 x}{7} - \frac{\cos^9 x}{9} + C$$

The final answer is $\boxed{-\frac{\cos^5 x}{5} + \frac{2\cos^7 x}{7} - \frac{\cos^9 x}{9} + C}$.