To solve the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \), we will follow these steps:
### Step 1: Identify the integral
We need to evaluate the integral:
\[
\int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt
\]
### Step 2: Use substitution
To solve this integral, we will use the substitution method. Let:
\[
u = 3t \quad \Rightarrow \quad du = 3 \, dt \quad \Rightarrow \quad dt = \frac{du}{3}
\]
### Step 3: Change the limits of integration
When \( t = 0 \):
\[
u = 3 \cdot 0 = 0
\]
When \( t = \frac{\pi}{2} \):
\[
u = 3 \cdot \frac{\pi}{2} = \frac{3\pi}{2}
\]
### Step 4: Rewrite the integral
Now we can rewrite the integral in terms of \( u \):
\[
\int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt = \int_{0}^{\frac{3\pi}{2}} \cos(u) \cdot \frac{du}{3}
\]
This simplifies to:
\[
\frac{1}{3} \int_{0}^{\frac{3\pi}{2}} \cos(u) \, du
\]
### Step 5: Evaluate the integral
The integral of \( \cos(u) \) is \( \sin(u) \):
\[
\frac{1}{3} \left[ \sin(u) \right]_{0}^{\frac{3\pi}{2}}
\]
Now we evaluate this at the limits:
\[
= \frac{1}{3} \left( \sin\left(\frac{3\pi}{2}\right) - \sin(0) \right)
\]
We know that:
\[
\sin\left(\frac{3\pi}{2}\right) = -1 \quad \text{and} \quad \sin(0) = 0
\]
Thus:
\[
= \frac{1}{3} \left( -1 - 0 \right) = -\frac{1}{3}
\]
### Final Answer
The value of the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \) is:
\[
-\frac{1}{3}
\]
To solve the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \), we will follow these steps:
### Step 1: Identify the integral
We need to evaluate the integral:
\[
\int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt
\]
...
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