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To solve the integral 

\[
\int_0^{\pi} \sin^2 t \cos^2 t \, dt,
\]

we can use a trigonometric identity to simplify it. The product \(\sin^2 t \cos^2 t\) can be rewritten using the identity:

\[
\sin^2 t \cos^2 t = \frac{1}{4} \sin^2(2t).
\]

Thus, the integral becomes:

\[
\int_0^{\pi} \sin^2 t \cos^2 t \, dt = \int_0^{\pi} \frac{1}{4} \sin^2(2t) \, dt = \frac{1}{4} \int_0^{\pi} \sin^2(2t) \, dt.
\]

Next, we use the identity for \(\sin^2(2t)\):

\[
\sin^2(2t) = \frac{1 - \cos(4t)}{2}.
\]

Substituting this, we get:

\[
\frac{1}{4} \int_0^{\pi} \sin^2(2t) \, dt = \frac{1}{4} \int_0^{\pi} \frac{1 - \cos(4t)}{2} \, dt = \frac{1}{8} \int_0^{\pi} (1 - \cos(4t)) \, dt.
\]

Now, we can split the integral:

\[
= \frac{1}{8} \left( \int_0^{\pi} 1 \, dt - \int_0^{\pi} \cos(4t) \, dt \right).
\]

The first part, \(\int_0^{\pi} 1 \, dt\), is simply \(\pi\), and the second part, \(\int_0^{\pi} \cos(4t) \, dt\), is zero because the integral of \(\cos(4t)\) over a full period is zero. Thus:

\[
\frac{1}{8} \left( \pi - 0 \right) = \frac{\pi}{8}.
\]

Therefore, the value of the integral is:

\[
\int_0^{\pi} \sin^2 t \cos^2 t \, dt = \frac{\pi}{8}.
\]

Would you like further explanation on any step?

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Here are some related questions to deepen understanding:

1. How can trigonometric identities simplify integrals involving products of sines and cosines?
2. What are other methods to integrate \(\sin^2 t \cos^2 t\) directly without identities?
3. How does symmetry help in evaluating integrals over symmetric intervals?
4. What is the significance of period integrals in trigonometric functions?
5. Can this approach be applied to different limits, like \([0, 2\pi]\)?

**Tip:** Remember that trigonometric identities are powerful tools for simplifying integrals with trigonometric functions.

Evaluate the integral from 0 to π of sin^2(t) cos^2(t) dt.

Explore the step-by-step solution for evaluating the integral of sin^2(t) cos^2(t) from 0 to π. This problem uses trigonometric identities to simplify the expression and involves calculus concepts. Ideal for undergraduate students in calculus.

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