Consider the integral `I_(m) = int_(0)^(pi) (sin2mx)/(sinx ) dx`, where m is a positive integer.
What is `I_(1)` equal to ?

Options

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

Explanation

To solve the integral \( I_1 = \int_0^{\pi} \frac{\sin 2x}{\sin x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I_1 = \int_0^{\pi} \frac{\sin 2x}{\sin x} \, dx \]
Using the double angle identity for sine, we have:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ I_1 = \int_0^{\pi} \frac{2 \sin x \cos x}{\sin x} \, dx \]

### Step 2: Simplify the integral
The \(\sin x\) in the numerator and denominator cancels out (for \(x \neq n\pi\), which is not an issue in the interval from \(0\) to \(\pi\)):
\[ I_1 = \int_0^{\pi} 2 \cos x \, dx \]

### Step 3: Integrate
Now, we can integrate \(2 \cos x\):
\[ I_1 = 2 \int_0^{\pi} \cos x \, dx \]
The integral of \(\cos x\) is \(\sin x\), so we have:
\[ I_1 = 2 [\sin x]_0^{\pi} \]
Calculating the limits:
\[ I_1 = 2 [\sin \pi - \sin 0] = 2 [0 - 0] = 2 \cdot 0 = 0 \]

### Final Answer
Thus, we conclude that:
\[ I_1 = 0 \]

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