`int sin^(3) x cos^(3) x dx =?`

Answer

To solve the integral \( \int \sin^3 x \cos^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int \sin^3 x \cos^3 x \, dx \]
We can express \( \sin^3 x \) as \( \sin x \cdot \sin^2 x \) and use the identity \( \sin^2 x = 1 - \cos^2 x \):
\[ \sin^3 x = \sin x (1 - \cos^2 x) \]
Thus, we can rewrite the integral as:
\[ \int \sin x (1 - \cos^2 x) \cos^3 x \, dx \]

### Step 2: Substitute \( t = \cos x \)
Let \( t = \cos x \). Then, the derivative \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \). Substituting these into the integral gives:
\[ \int \sin x (1 - \cos^2 x) \cos^3 x \, dx = \int (1 - t^2) t^3 (-dt) \]
This simplifies to:
\[ -\int (1 - t^2) t^3 \, dt \]

### Step 3: Expand the integrand
Now, we expand the integrand:
\[ -\int (t^3 - t^5) \, dt \]

### Step 4: Integrate term by term
Now we can integrate term by term:
\[ -\left( \frac{t^4}{4} - \frac{t^6}{6} \right) + C \]
This simplifies to:
\[ -\frac{t^4}{4} + \frac{t^6}{6} + C \]

### Step 5: Substitute back for \( t \)
Now, we substitute back \( t = \cos x \):
\[ -\frac{\cos^4 x}{4} + \frac{\cos^6 x}{6} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \sin^3 x \cos^3 x \, dx = \frac{\cos^6 x}{6} - \frac{\cos^4 x}{4} + C \]

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