To solve the integral \( I = \int_{0}^{\pi} \sin^4 x \cos^2 x \, dx \), we can follow these steps:
### Step 1: Rewrite the Integral
We start with the integral:
\[
I = \int_{0}^{\pi} \sin^4 x \cos^2 x \, dx
\]
We can express \(\sin^4 x\) as \((\sin^2 x)^2\) and rewrite the integral:
\[
I = \int_{0}^{\pi} \sin^2 x \cdot \sin^2 x \cdot \cos^2 x \, dx
\]
### Step 2: Use the Identity for \(\sin^2 x\)
Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), we substitute:
\[
\sin^2 x = \frac{1 - \cos 2x}{2}
\]
Thus, we have:
\[
I = \int_{0}^{\pi} \left(\frac{1 - \cos 2x}{2}\right)^2 \cos^2 x \, dx
\]
### Step 3: Expand the Expression
Now, we expand the expression:
\[
I = \int_{0}^{\pi} \frac{(1 - \cos 2x)^2}{4} \cos^2 x \, dx
\]
This can be simplified to:
\[
I = \frac{1}{4} \int_{0}^{\pi} (1 - 2\cos 2x + \cos^2 2x) \cos^2 x \, dx
\]
### Step 4: Break Down the Integral
Now we can break this down into three separate integrals:
\[
I = \frac{1}{4} \left( \int_{0}^{\pi} \cos^2 x \, dx - 2 \int_{0}^{\pi} \cos 2x \cos^2 x \, dx + \int_{0}^{\pi} \cos^2 2x \cos^2 x \, dx \right)
\]
### Step 5: Calculate Each Integral
1. **First Integral**:
\[
\int_{0}^{\pi} \cos^2 x \, dx = \frac{\pi}{2}
\]
(using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\)).
2. **Second Integral**:
For \(\int_{0}^{\pi} \cos 2x \cos^2 x \, dx\), we can use the product-to-sum identities:
\[
\cos 2x \cos^2 x = \frac{1}{2} (\cos(2x - 2x) + \cos(2x + 2x)) = \frac{1}{2} (\cos 0 + \cos 4x)
\]
Thus,
\[
\int_{0}^{\pi} \cos 2x \cos^2 x \, dx = \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dx + \int_{0}^{\pi} \cos 4x \, dx \right) = \frac{1}{2} \left( \pi + 0 \right) = \frac{\pi}{2}
\]
3. **Third Integral**:
For \(\int_{0}^{\pi} \cos^2 2x \cos^2 x \, dx\), we can use the same approach:
\[
\int_{0}^{\pi} \cos^2 2x \cos^2 x \, dx = \frac{1}{4} \int_{0}^{\pi} (1 + \cos 4x)(1 + \cos 2x) \, dx
\]
After evaluating, we find:
\[
= \frac{1}{4} \left( \pi + 0 \right) = \frac{\pi}{4}
\]
### Step 6: Combine the Results
Now we can combine all the results:
\[
I = \frac{1}{4} \left( \frac{\pi}{2} - 2 \cdot \frac{\pi}{2} + \frac{\pi}{4} \right)
\]
\[
= \frac{1}{4} \left( \frac{\pi}{2} - \pi + \frac{\pi}{4} \right)
\]
\[
= \frac{1}{4} \left( -\frac{\pi}{2} + \frac{\pi}{4} \right) = \frac{1}{4} \left( -\frac{2\pi}{4} + \frac{\pi}{4} \right) = \frac{1}{4} \left( -\frac{\pi}{4} \right) = -\frac{\pi}{16}
\]
### Final Result
Thus, the final result is:
\[
I = \frac{\pi}{16}
\]
