To integrate the function $-2x \sin^2(x)$, we will apply the following steps:

Step 1: Use the trigonometric identity for $\sin^2(x)$

We can rewrite $\sin^2(x)$ using the identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ So, the original integral becomes: $\int -2x \sin^2(x) \, dx = \int -2x \cdot \frac{1 - \cos(2x)}{2} \, dx$ Simplifying: $\int -2x \sin^2(x) \, dx = \int -x (1 - \cos(2x)) \, dx$

Step 2: Break the integral into two parts

Now, split the integral: $\int -x(1 - \cos(2x)) \, dx = \int -x \, dx + \int x \cos(2x) \, dx$

Step 3: Solve each part

1. First Integral: $\int -x \, dx$ $\int -x \, dx = -\frac{x^2}{2}$

2. Second Integral: $\int x \cos(2x) \, dx$

This requires integration by parts. Let:

• $u = x$, so $du = dx$
• $dv = \cos(2x) dx$, so $v = \frac{\sin(2x)}{2}$

Now, apply the integration by parts formula $\int u dv = uv - \int v du$: $\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} - \int \frac{\sin(2x)}{2} dx$ The remaining integral is: $\int \frac{\sin(2x)}{2} \, dx = -\frac{\cos(2x)}{4}$ Thus: $\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4}$

Step 4: Combine results

Now, putting everything together: $\int -2x \sin^2(x) \, dx = -\frac{x^2}{2} + \left( \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} \right) + C$ where $C$ is the constant of integration.

Thus, the final result is: $\boxed{-\frac{x^2}{2} + \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} + C}$

Would you like further clarification or more details on any step?

Relative Questions:

1. How is the trigonometric identity for $\sin^2(x)$ derived?
2. Why is integration by parts necessary for $x \cos(2x)$?
3. Can integration by parts be applied in different orders for different results?
4. How do we simplify $\int \sin(2x) \, dx$?
5. What are the general strategies for integrating products of trigonometric and polynomial functions?

Tip:

When dealing with trigonometric integrals, using identities to simplify the expression before integration can often make the process easier.