Consider the integral `I_(m) = int_(0)^(pi) (sin2mx)/(sinx ) dx`, where m is a positive integer.
What is `I_(2) + I_(3)` equal to ?

Options

A. `4`
B. `2`
C. `1`
D. `0`

Explanation

To solve the integral \( I_m = \int_0^{\pi} \frac{\sin(2mx)}{\sin(x)} \, dx \) for \( m = 2 \) and \( m = 3 \), we will calculate \( I_2 + I_3 \).

### Step 1: Calculate \( I_2 \)

We start with:
\[ I_2 = \int_0^{\pi} \frac{\sin(4x)}{\sin(x)} \, dx \]
Using the identity \( \sin(2a) = 2 \sin(a) \cos(a) \), we can express \( \sin(4x) \) as:
\[ \sin(4x) = 2 \sin(2x) \cos(2x) = 2(2 \sin(x) \cos(x)) \cos(2x) = 4 \sin(x) \cos(x) \cos(2x) \]
Thus:
\[ I_2 = \int_0^{\pi} 4 \cos(x) \cos(2x) \, dx \]

### Step 2: Simplify \( I_2 \)

Now we can separate the integral:
\[ I_2 = 4 \int_0^{\pi} \cos(x) \cos(2x) \, dx \]
Using the product-to-sum identities:
\[ \cos(x) \cos(2x) = \frac{1}{2}(\cos(x + 2x) + \cos(x - 2x)) = \frac{1}{2}(\cos(3x) + \cos(-x)) = \frac{1}{2}(\cos(3x) + \cos(x)) \]
Thus:
\[ I_2 = 4 \int_0^{\pi} \frac{1}{2}(\cos(3x) + \cos(x)) \, dx = 2 \left( \int_0^{\pi} \cos(3x) \, dx + \int_0^{\pi} \cos(x) \, dx \right) \]

### Step 3: Evaluate the integrals

The integral of \( \cos(3x) \) over the interval \( [0, \pi] \):
\[ \int_0^{\pi} \cos(3x) \, dx = \left[ \frac{\sin(3x)}{3} \right]_0^{\pi} = \frac{\sin(3\pi)}{3} - \frac{\sin(0)}{3} = 0 \]
The integral of \( \cos(x) \) over the interval \( [0, \pi] \):
\[ \int_0^{\pi} \cos(x) \, dx = \left[ \sin(x) \right]_0^{\pi} = \sin(\pi) - \sin(0) = 0 \]
Thus:
\[ I_2 = 2(0 + 0) = 0 \]

### Step 4: Calculate \( I_3 \)

Now we calculate \( I_3 \):
\[ I_3 = \int_0^{\pi} \frac{\sin(6x)}{\sin(x)} \, dx \]
Using the same approach:
\[ I_3 = \int_0^{\pi} 6 \cos(3x) \cos(3x) \, dx \]
Using the product-to-sum identities:
\[ \cos(3x) \cos(3x) = \frac{1}{2}(1 + \cos(6x)) \]
Thus:
\[ I_3 = 6 \int_0^{\pi} \frac{1}{2}(1 + \cos(6x)) \, dx = 3 \left( \int_0^{\pi} 1 \, dx + \int_0^{\pi} \cos(6x) \, dx \right) \]
The integral of \( 1 \) over \( [0, \pi] \):
\[ \int_0^{\pi} 1 \, dx = \pi \]
The integral of \( \cos(6x) \) over \( [0, \pi] \):
\[ \int_0^{\pi} \cos(6x) \, dx = \left[ \frac{\sin(6x)}{6} \right]_0^{\pi} = \frac{\sin(6\pi)}{6} - \frac{\sin(0)}{6} = 0 \]
Thus:
\[ I_3 = 3(\pi + 0) = 3\pi \]

### Step 5: Combine \( I_2 \) and \( I_3 \)

Finally, we combine the results:
\[ I_2 + I_3 = 0 + 3\pi = 3\pi \]

### Final Answer
\[ I_2 + I_3 = 3\pi \]