Evaluate `intsin^(3)x cos^(2)x dx`

Explanation

To evaluate the integral \(\int \sin^3 x \cos^2 x \, dx\), we can follow these steps:

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

### Step 2: Substitute for \(u\)
Let \(u = \cos x\). Then, the derivative \(du = -\sin x \, dx\) or \(-du = \sin x \, dx\). We can rewrite the integral in terms of \(u\):
\[ \int (1 - u^2) u^2 (-du) = -\int (1 - u^2) u^2 \, du \]

### Step 3: Expand the integrand
Now, expand the integrand:
\[ -\int (u^2 - u^4) \, du = -\left( \int u^2 \, du - \int u^4 \, du \right) \]

### Step 4: Integrate term by term
Now we can integrate each term:
\[ -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = -\frac{u^3}{3} + \frac{u^5}{5} + C \]

### Step 5: Substitute back for \(u\)
Now, substitute back \(u = \cos x\):
\[ -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \sin^3 x \cos^2 x \, dx = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \]

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