To evaluate the integral \( I = \int_0^{\pi} \sin^4 x \cos^2 x \, dx \), we can follow these steps:
### Step 1: Use the property of definite integrals
We can use the property of definite integrals:
\[
\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx
\]
In our case, let \( a = \pi/2 \) and \( f(x) = \sin^4 x \cos^2 x \). Then,
\[
I = \int_0^{\pi} \sin^4 x \cos^2 x \, dx = \int_0^{\pi/2} \sin^4 x \cos^2 x \, dx + \int_0^{\pi/2} \sin^4(\pi - x) \cos^2(\pi - x) \, dx
\]
Since \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we have:
\[
I = 2 \int_0^{\pi/2} \sin^4 x \cos^2 x \, dx
\]
### Step 2: Substitute \( \cos^2 x \)
Next, we can express \( \cos^2 x \) in terms of \( \sin^2 x \):
\[
\cos^2 x = 1 - \sin^2 x
\]
Thus, we rewrite the integral:
\[
I = 2 \int_0^{\pi/2} \sin^4 x (1 - \sin^2 x) \, dx
\]
This simplifies to:
\[
I = 2 \left( \int_0^{\pi/2} \sin^4 x \, dx - \int_0^{\pi/2} \sin^6 x \, dx \right)
\]
### Step 3: Use the reduction formula
We will use the reduction formula for integrals of the form \( \int_0^{\pi/2} \sin^n x \, dx \):
\[
G_n = \int_0^{\pi/2} \sin^n x \, dx = \frac{n-1}{n} G_{n-2}
\]
Calculating \( G_4 \) and \( G_6 \):
1. For \( G_4 \):
\[
G_4 = \frac{3}{4} G_2
\]
And \( G_2 = \int_0^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4} \), so:
\[
G_4 = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}
\]
2. For \( G_6 \):
\[
G_6 = \frac{5}{6} G_4 = \frac{5}{6} \cdot \frac{3\pi}{16} = \frac{15\pi}{96} = \frac{5\pi}{32}
\]
### Step 4: Substitute back into the equation for \( I \)
Now we substitute \( G_4 \) and \( G_6 \) back into the expression for \( I \):
\[
I = 2 \left( G_4 - G_6 \right) = 2 \left( \frac{3\pi}{16} - \frac{5\pi}{32} \right)
\]
To combine these, we need a common denominator:
\[
\frac{3\pi}{16} = \frac{6\pi}{32}
\]
Thus:
\[
I = 2 \left( \frac{6\pi}{32} - \frac{5\pi}{32} \right) = 2 \left( \frac{\pi}{32} \right) = \frac{\pi}{16}
\]
### Final Answer
The value of the integral is:
\[
\boxed{\frac{\pi}{16}}
\]
