The integral you're asking for is:

\[
I = \int_0^t \sin(kx) \, dx
\]

To solve this, we can integrate \(\sin(kx)\) with respect to \(x\).

### Step 1: Find the antiderivative of \(\sin(kx)\)
We know that:

\[
\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx)
\]

### Step 2: Apply the limits of integration (from 0 to \(t\))

Now, applying the limits to the antiderivative:

\[
I = \left[-\frac{1}{k} \cos(kx)\right]_0^t
\]

### Step 3: Substitute the limits

Substitute \(x = t\) and \(x = 0\):

\[
I = -\frac{1}{k} \cos(kt) + \frac{1}{k} \cos(0)
\]

Since \(\cos(0) = 1\), this becomes:

\[
I = -\frac{1}{k} \cos(kt) + \frac{1}{k}
\]

### Final Answer:

\[
I = \frac{1}{k} (1 - \cos(kt))
\]

Would you like more details on any of the steps?

Here are five related questions to deepen your understanding:
1. How would the integral change if you replaced \(\sin(kx)\) with \(\cos(kx)\)?
2. What if the limits of integration were from \(a\) to \(b\) instead of from 0 to \(t\)?
3. Can you evaluate the integral for specific values of \(k\) and \(t\)?
4. How would you approach the integration of \(\sin^2(kx)\) from 0 to \(t\)?
5. How does the result relate to the average value of \(\sin(kx)\) over the interval \([0, t]\)?

**Tip:** When integrating trigonometric functions, it’s often helpful to remember the standard antiderivatives of \(\sin(x)\) and \(\cos(x)\).

Explore the step-by-step solution to the integral of sin(kx) from 0 to t. This problem covers key concepts in calculus, including integration of trigonometric functions and the Fundamental Theorem of Calculus. Ideal for students in Grades 11-12 or early college.

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