To integrate $I = \int x \cos(x) \sin(nx) \, dx$, we'll use the integration by parts repeatedly, following the formula you mentioned:

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

We define our components as:

• $u = x$ (since it simplifies with differentiation),
• $dv = \cos(x) \sin(nx) \, dx$.

Now, we need to repeatedly differentiate $u$ and integrate $dv$ three times, applying the following steps:

Step 1: First Integration by Parts

Let $u = x$ and $dv = \cos(x) \sin(nx) \, dx$.

1. Differentiate $u$:
   $du = dx$

2. Integrate $dv$ (let's use a trigonometric identity):
   $\cos(x) \sin(nx) = \frac{1}{2} [\sin((n+1)x) - \sin((n-1)x)]$

   So, $dv = \cos(x) \sin(nx) \, dx = \frac{1}{2} \left[\sin((n+1)x) - \sin((n-1)x)\right] dx$

   After integrating: [ v = \frac{1}{2(n+1)} \left[ \cos((n+1)x) \right] - \frac{1}{2(n-1)} \left[ \cos((n-1)x) \right] ]

3. Now apply the formula: $\int x \cos(x) \sin(nx) \, dx = x \left( \frac{1}{2(n+1)} \cos((n+1)x) - \frac{1}{2(n-1)} \cos((n-1)x) \right) - \int \left( \frac{1}{2(n+1)} \cos((n+1)x) - \frac{1}{2(n-1)} \cos((n-1)x) \right) \, dx$

Step 2: Solve the remaining integral

The remaining integral involves two simpler integrals: $\int \cos((n+1)x) \, dx = \frac{1}{n+1} \sin((n+1)x)$ and $\int \cos((n-1)x) \, dx = \frac{1}{n-1} \sin((n-1)x).$

Final Answer

Putting everything together, we get: $I = \frac{x}{2(n+1)} \cos((n+1)x) - \frac{x}{2(n-1)} \cos((n-1)x) + \frac{1}{2(n+1)^2} \sin((n+1)x) - \frac{1}{2(n-1)^2} \sin((n-1)x) + C$

This is the integrated result for $\int x \cos(x) \sin(nx) \, dx$.

Would you like more details or have any questions? Here are five related questions to deepen understanding:

1. How does integration by parts work for non-trigonometric functions?
2. What are other methods to integrate trigonometric functions?
3. How do you handle integrals involving products of trigonometric functions?
4. Can you apply integration by parts to solve a double integral?
5. How would the integral change if the product involved $\cos^2(x)$ or $\sin^2(nx)$?

Tip: Whenever integrating products of functions, carefully choose $u$ and $dv$ to simplify the overall process.