To integrate $\sin^5(x) \cdot \cos^4(x) \, dx$, we can use the following approach:

Step 1: Express the sine and cosine powers in terms of sine and cosine functions

We will separate one $\sin(x)$ and one $\cos(x)$ from the original integral to make the substitution easier:

$\sin^5(x) \cdot \cos^4(x) = \sin^4(x) \cdot \sin(x) \cdot \cos^4(x)$

This can be written as:

$\sin^4(x) \cdot \cos^4(x) \cdot \sin(x) \, dx$

Step 2: Use trigonometric identities

We can rewrite $\sin^4(x)$ as $(\sin^2(x))^2$ and $\cos^4(x)$ as $(\cos^2(x))^2$:

$(\sin^2(x))^2 \cdot (\cos^2(x))^2 \cdot \sin(x) \, dx$

Now, recall that $\sin^2(x) = 1 - \cos^2(x)$. Let's make the substitution $u = \cos(x)$, so that $du = -\sin(x) \, dx$. The integral now becomes:

$\int (1 - u^2)^2 \cdot u^4 \cdot (-du)$

Step 3: Simplify and expand the expression

Expanding $(1 - u^2)^2$ gives:

$(1 - u^2)^2 = 1 - 2u^2 + u^4$

So the integral becomes:

$-\int (1 - 2u^2 + u^4) \cdot u^4 \, du$

Expanding this:

$-\int \left( u^4 - 2u^6 + u^8 \right) \, du$

Step 4: Integrate term by term

Now integrate each term separately:

$-\left[ \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} \right] + C$

Finally, substitute back $u = \cos(x)$:

[ -\left[ \frac{\cos^5(x)}