`int cos^(3) x. sin x dx`

Answer

To solve the integral \( \int \cos^3 x \sin x \, dx \), we can use the substitution method. Here are the steps:

### Step 1: Choose a substitution
Let \( t = \cos x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = -\sin x \quad \Rightarrow \quad dt = -\sin x \, dx \]
This implies that \( \sin x \, dx = -dt \).

### Step 2: Rewrite the integral
Substituting \( t \) and \( dt \) into the integral, we have:
\[ \int \cos^3 x \sin x \, dx = \int t^3 (-dt) = -\int t^3 \, dt \]

### Step 3: Integrate
Now, we can integrate \( -\int t^3 \, dt \):
\[ -\int t^3 \, dt = -\left( \frac{t^4}{4} \right) + C = -\frac{t^4}{4} + C \]

### Step 4: Substitute back for \( t \)
Now, substitute back \( t = \cos x \):
\[ -\frac{t^4}{4} + C = -\frac{\cos^4 x}{4} + C \]

### Step 5: Final answer
Thus, the integral \( \int \cos^3 x \sin x \, dx \) is:
\[ -\frac{\cos^4 x}{4} + C \]

### Summary of the solution:
\[ \int \cos^3 x \sin x \, dx = -\frac{\cos^4 x}{4} + C \]