The integral of \(\sin(4x)\) with respect to \(x\) can be solved using basic integration rules. Here's the solution:

\[
\int \sin(4x) \, dx
\]

To solve this, we use substitution. Let:

\[
u = 4x \quad \text{which means} \quad du = 4 \, dx \quad \text{or} \quad dx = \frac{du}{4}
\]

Now substitute \(u\) into the integral:

\[
\int \sin(4x) \, dx = \int \sin(u) \cdot \frac{du}{4}
\]

Factor out the constant \(\frac{1}{4}\):

\[
\frac{1}{4} \int \sin(u) \, du
\]

The integral of \(\sin(u)\) is \(-\cos(u)\), so:

\[
\frac{1}{4} \cdot (-\cos(u)) + C
\]

Substitute \(u = 4x\) back into the equation:

\[
-\frac{1}{4} \cos(4x) + C
\]

Thus, the integral is:

\[
\int \sin(4x) \, dx = -\frac{1}{4} \cos(4x) + C
\]

### Would you like more details on the steps or have any questions?

Here are 5 related questions:
1. How do you integrate functions with other trigonometric identities, such as \(\cos(5x)\)?
2. What is the integral of a product of sine and cosine, such as \(\sin(x)\cos(x)\)?
3. Can substitution be used to solve integrals of higher powers of trigonometric functions?
4. What is the derivative of \(-\frac{1}{4} \cos(4x)\)?
5. How does the integral of \(\sin(4x)\) compare to \(\sin(kx)\), where \(k\) is a constant?

**Tip**: Always look for substitutions in integrals involving complex trigonometric functions; it often simplifies the process significantly.

Solve the integral of sin(4x) using substitution with detailed steps. Learn how to simplify trigonometric integrals with substitution techniques. Suitable for Grades 11-12 and college-level calculus students.

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