$$ \int_{}^{}\!\sin\,x\,\cos\,x\,\mathrm{d}x $$

To integrate $\int \sin x \cos x \, dx$, we can use a trigonometric identity to simplify the integrand. One useful identity is:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]

Using this identity, the integral becomes:
\[ \int \sin x \cos x \, dx = \int \frac{1}{2} \sin 2x \, dx \]

We can factor out the constant $\frac{1}{2}$:
\[ \int \frac{1}{2} \sin 2x \, dx = \frac{1}{2} \int \sin 2x \, dx \]

Next, we integrate $\sin 2x$. The integral of $\sin u$ is $-\cos u$, where $u = 2x$:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \]

Combining these results, we have:
\[ \frac{1}{2} \int \sin 2x \, dx = \frac{1}{2} \left(-\frac{1}{2} \cos 2x\right) = -\frac{1}{4} \cos 2x + C \]

Therefore, the integral is:
\[ \int \sin x \cos x \, dx = -\frac{1}{4} \cos 2x + C \]

Would you like more details or have any questions about this solution?

Here are 5 related questions you can ask:
1. What is the integral of $\sin^2 x$?
2. How do you integrate $\cos^2 x$?
3. Can you explain how to use integration by parts?
4. How do you find the derivative of $\cos x$?
5. What are some techniques for integrating trigonometric functions?

**Tip:** Recognizing and using trigonometric identities can simplify the process of integrating trigonometric functions.

Learn how to integrate $ \sin x \cos x $ step-by-step using trigonometric identities and integration techniques. This problem involves calculus and trigonometry concepts and is suitable for advanced high school students and beyond.

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