The exercise asks to solve the integral:

$\int x^2 \cos x \, dx$

Solution:

This integral is best solved using integration by parts, where:

$\int u \, dv = uv - \int v \, du$

Step 1: Assign $u$ and $dv$

Let: $u = x^2 \quad \text{and} \quad dv = \cos x \, dx$

Step 2: Compute $du$ and $v$

• $du = 2x \, dx$
• $v = \int \cos x \, dx = \sin x$

Step 3: Apply the formula

Using the integration by parts formula: $\int x^2 \cos x \, dx = u v - \int v \, du$ Substitute: $\int x^2 \cos x \, dx = x^2 \sin x - \int \sin x (2x \, dx)$

Step 4: Solve the remaining integral

The new integral is: $\int 2x \sin x \, dx$ We apply integration by parts again. Let: $u = 2x \quad \text{and} \quad dv = \sin x \, dx$ Then: $du = 2 \, dx \quad \text{and} \quad v = -\cos x$

Using the formula: $\int 2x \sin x \, dx = -2x \cos x + \int 2 \cos x \, dx$

Step 5: Solve $\int 2 \cos x \, dx$

$\int 2 \cos x \, dx = 2 \sin x$

Step 6: Combine results

$\int 2x \sin x \, dx = -2x \cos x + 2 \sin x$

Step 7: Substitute back

Returning to the original expression: $\int x^2 \cos x \, dx = x^2 \sin x - \left(-2x \cos x + 2 \sin x\right)$ Simplify: $\int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + C$

Final Answer:

$\boxed{\int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x + C}$

Let me know if you’d like further clarification or details!

Here are five related questions to expand your understanding:

1. How is the integration by parts formula derived?
2. What are common scenarios where integration by parts is most useful?
3. How can we solve $\int x^n e^x \, dx$ using a similar approach?
4. What is the significance of the constant of integration, $C$, in indefinite integrals?
5. Can we use substitution instead of integration by parts for this problem? Why or why not?

Tip: When solving integration problems, carefully choose $u$ and $dv$ to simplify the process!