https://file.aiquickdraw.com/mathbot/file/1729151926138yk3jdyem.jpg

The integral given in the image is:

\[
\int -6x \sin(3x) \, dx
\]

To solve this, we can use **integration by parts**. For an integral of the form:

\[
\int u \, dv = uv - \int v \, du
\]

We need to choose \( u \) and \( dv \). Let:

- \( u = -6x \), so \( du = -6 \, dx \)
- \( dv = \sin(3x) \, dx \), so \( v = -\frac{1}{3} \cos(3x) \) (since the integral of \( \sin(3x) \) is \( -\frac{1}{3} \cos(3x) \)).

Now, applying the formula for integration by parts:

\[
\int -6x \sin(3x) \, dx = u v - \int v \, du
\]

Substitute \( u \), \( v \), and \( du \):

\[
= (-6x) \left( -\frac{1}{3} \cos(3x) \right) - \int \left( -\frac{1}{3} \cos(3x) \right) (-6) \, dx
\]

Simplify:

\[
= 2x \cos(3x) - 2 \int \cos(3x) \, dx
\]

The integral of \( \cos(3x) \) is \( \frac{1}{3} \sin(3x) \). Therefore:

\[
2 \int \cos(3x) \, dx = 2 \cdot \frac{1}{3} \sin(3x) = \frac{2}{3} \sin(3x)
\]

Thus, the final solution is:

\[
2x \cos(3x) - \frac{2}{3} \sin(3x) + C
\]

where \( C \) is the constant of integration.

Would you like any further details or have questions on this process?

Here are 5 questions related to this topic:
1. How do you determine which part to assign as \( u \) and \( dv \) in integration by parts?
2. What is the general rule for integrating products of functions like \( x \sin(x) \)?
3. Can integration by parts be applied multiple times if necessary?
4. How do you solve integrals that involve both trigonometric and polynomial terms?
5. What other methods besides integration by parts could work for certain products of functions?

**Tip:** When using integration by parts, try to choose \( u \) as a function that simplifies upon differentiation (like \( x \), \( x^2 \), etc.).

Evaluate the integral: ∫ -6x sin(3x) dx

Solve the integral ∫ -6x sin(3x) dx step-by-step using integration by parts. Learn how polynomial and trigonometric functions are integrated together with detailed explanations.

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