To solve the integral

$\int x \sin^2(x) \, dx,$

we can proceed by using integration by parts and a trigonometric identity for $\sin^2(x)$.

Step 1: Rewrite $\sin^2(x)$ using a trigonometric identity

The identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ allows us to rewrite the integrand as: $\int x \sin^2(x) \, dx = \int x \cdot \frac{1 - \cos(2x)}{2} \, dx.$ This simplifies to: $\int x \sin^2(x) \, dx = \frac{1}{2} \int x (1 - \cos(2x)) \, dx.$

Expanding this, we get: $\frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos(2x) \, dx.$

Step 2: Solve each integral separately

Integral 1: $\frac{1}{2} \int x \, dx$

This is straightforward: $\frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}.$

Integral 2: $-\frac{1}{2} \int x \cos(2x) \, dx$

For this integral, we can use integration by parts. Let $u = x$ and $dv = \cos(2x) \, dx$.

Then, $du = dx$ and we integrate $dv$ to get $v = \frac{\sin(2x)}{2}$.

Using integration by parts, we have: $\int x \cos(2x) \, dx = x \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx.$

Simplifying: $= \frac{x \sin(2x)}{2} - \frac{1}{2} \int \sin(2x) \, dx.$

The integral of $\sin(2x)$ is $-\frac{\cos(2x)}{2}$, so we get: $= \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4}.$

Putting it all together

Substitute back into the original expression: $\int x \sin^2(x) \, dx = \frac{x^2}{4} - \frac{1}{2} \left( \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} \right).$

Simplify: $= \frac{x^2}{4} - \frac{x \sin(2x)}{4} - \frac{\cos(2x)}{8} + C,$ where $C$ is the constant of integration.

So, the final answer is: $\int x \sin^2(x) \, dx = \frac{x^2}{4} - \frac{x \sin(2x)}{4} - \frac{\cos(2x)}{8} + C.$

Would you like more details on any specific step?

Here are some related questions to deepen understanding:

1. How can we use the integration by parts method for different trigonometric integrals?
2. What are other trigonometric identities useful in integrals?
3. How does substitution help in solving integrals like $\int x \sin^2(x) \, dx$?
4. Can this approach be extended to integrals involving $\sin^3(x)$ or $\sin^4(x)$?
5. Why is integration by parts particularly useful in integrals involving products of functions?

Tip: Recognizing when to apply trigonometric identities can simplify many integrals, making them more manageable.