To compute the integral of \( -2x \sin^2(x) \, dx \), we can follow these steps:

### Step 1: Use a trigonometric identity
Recall the identity for \( \sin^2(x) \):
\[
\sin^2(x) = \frac{1 - \cos(2x)}{2}
\]
Substituting this into the integral:

\[
\int -2x \sin^2(x) \, dx = \int -2x \cdot \frac{1 - \cos(2x)}{2} \, dx
\]
Simplifying:
\[
\int -2x \sin^2(x) \, dx = \int -x(1 - \cos(2x)) \, dx
\]
This gives:
\[
\int -x(1 - \cos(2x)) \, dx = \int -x \, dx + \int x \cos(2x) \, dx
\]

### Step 2: Solve the first integral
\[
\int -x \, dx = -\frac{x^2}{2}
\]

### Step 3: Solve the second integral using integration by parts
Let \( u = x \) and \( dv = \cos(2x) \, dx \).

- \( du = dx \)
- \( v = \frac{\sin(2x)}{2} \) (since the integral of \( \cos(2x) \) is \( \frac{\sin(2x)}{2} \))

Now apply integration by parts:
\[
\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx
\]
\[
\int \frac{\sin(2x)}{2} \, dx = -\frac{\cos(2x)}{4}
\]
Thus:
\[
\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4}
\]

### Step 4: Combine the results
Now combine the two integrals:
\[
\int -2x \sin^2(x) \, dx = -\frac{x^2}{2} + \left( \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} \right) + C
\]

So the final answer is:
\[
\boxed{-\frac{x^2}{2} + \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} + C}
\]

Would you like more details on any step, or do you have any questions?

Here are 5 related questions:
1. How do you perform integration by parts for other trigonometric integrals?
2. Can you solve the integral of \( \sin^2(x) \) without substitution?
3. What is the integral of \( \cos^2(x) \)?
4. How does integration by parts work for multiple variables?
5. What are the common trigonometric identities used in integrals?

**Tip:** Always look for trigonometric identities when integrating powers of sine or cosine functions to simplify the process.

Learn how to compute the integral of -2x sin^2(x) dx using trigonometric identities and integration by parts. This solution covers essential calculus techniques for solving trigonometric integrals. Suitable for undergraduate students.

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