To solve the integral \( \int_{0}^{100\pi} \left( \sum_{r=1}^{10} \tan(rx) \right) dx \), we will follow these steps:
### Step 1: Define the function
Let \( f(x) = \sum_{r=1}^{10} \tan(rx) \).
This means we are summing the tangent functions for \( r = 1 \) to \( r = 10 \).
**Hint:** Recognize that \( \tan(rx) \) is a periodic function.
### Step 2: Identify the period of the function
The function \( \tan(rx) \) has a period of \( \frac{\pi}{r} \). Therefore, the function \( f(x) \) will have a period of \( \pi \) since the highest frequency term is \( \tan(10x) \).
**Hint:** The period of a sum of periodic functions is determined by the least common multiple of their individual periods.
### Step 3: Use the periodicity property of integrals
Since \( f(x) \) is periodic with period \( \pi \), we can use the property of integrals of periodic functions:
\[
\int_{0}^{a + nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx
\]
where \( T \) is the period of the function.
In our case, we have:
\[
\int_{0}^{100\pi} f(x) \, dx = 100 \int_{0}^{\pi} f(x) \, dx
\]
**Hint:** Break the integral over the larger interval into multiple integrals over one period.
### Step 4: Evaluate the integral over one period
Now we need to compute:
\[
\int_{0}^{\pi} f(x) \, dx = \int_{0}^{\pi} \left( \sum_{r=1}^{10} \tan(rx) \right) dx = \sum_{r=1}^{10} \int_{0}^{\pi} \tan(rx) \, dx
\]
**Hint:** You can interchange the summation and the integral due to linearity.
### Step 5: Evaluate \( \int_{0}^{\pi} \tan(rx) \, dx \)
Using the property of definite integrals:
\[
\int_{0}^{\pi} \tan(rx) \, dx = \int_{0}^{\pi} \tan(r(\pi - x)) \, dx = -\int_{0}^{\pi} \tan(rx) \, dx
\]
This implies:
\[
2 \int_{0}^{\pi} \tan(rx) \, dx = 0 \Rightarrow \int_{0}^{\pi} \tan(rx) \, dx = 0
\]
**Hint:** Use symmetry properties of the tangent function.
### Step 6: Conclude the integral evaluation
Since each integral \( \int_{0}^{\pi} \tan(rx) \, dx = 0 \), we have:
\[
\int_{0}^{\pi} f(x) \, dx = \sum_{r=1}^{10} 0 = 0
\]
Thus,
\[
\int_{0}^{100\pi} f(x) \, dx = 100 \cdot 0 = 0
\]
### Final Answer
The value of the integral \( \int_{0}^{100\pi} \left( \sum_{r=1}^{10} \tan(rx) \right) dx \) is \( \boxed{0} \).
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