Let's analyze the given integral problem and find the value of \({\rm{I}}_1\).

The integral is defined as \({\rm{I}}_{\rm{m}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}},\) where m is a positive integer.

We are asked to find the value of \({\rm{I}}_1\). This means we need to substitute \({\rm{m}} = 1\) into the integral definition.

Calculating the Integral I₁

To find \({\rm{I}}_1\), we set \({\rm{m}} = 1\) in the expression for \({\rm{I}}_{\rm{m}}\):

\[{\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin (2 \cdot 1 \cdot {\rm{x}})}}{{\sin {\rm{x}}}}{\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\]

Now, we need to evaluate this definite integral. We can use a trigonometric identity to simplify the integrand \(\frac{{\sin 2{\rm{x}}}}{{\sin {\rm{x}}}}\).

Recall the double angle identity for sine:

\[\sin 2\theta = 2 \sin \theta \cos \theta\]

Using this identity with \(\theta = {\rm{x}}\), we have \(\sin 2{\rm{x}} = 2 \sin {\rm{x}} \cos {\rm{x}}\).

Substitute this into the integral expression for \({\rm{I}}_1\):

\[{\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{2 \sin {\rm{x}} \cos {\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\]

For the interval of integration \(0 \le {\rm{x}} \le \pi\), \(\sin {\rm{x}}\) is zero only at the endpoints \({\rm{x}} = 0\) and \({\rm{x}} = \pi\). For \(0 < {\rm{x}} < \pi\), \(\sin {\rm{x}} > 0\). We can cancel the \(\sin {\rm{x}}\) term from the numerator and denominator for \(0 < {\rm{x}} < \pi\).

\[{\rm{I}}_1 = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos {\rm{x}} {\rm{dx}}\]

Now, we can evaluate this simple integral. The antiderivative of \(\cos {\rm{x}}\) is \(\sin {\rm{x}}\).

\[{\rm{I}}_1 = 2 \left[ \sin {\rm{x}} \right]_0^{\rm{\pi }}\]

Apply the limits of integration:

\[{\rm{I}}_1 = 2 (\sin \pi - \sin 0)\]

We know that \(\sin \pi = 0\) and \(\sin 0 = 0\).

\[{\rm{I}}_1 = 2 (0 - 0)\]

\[{\rm{I}}_1 = 2 \cdot 0\]

\[{\rm{I}}_1 = 0\]

Therefore, the value of \({\rm{I}}_1\) is 0.

Summary of Steps

• Identify the specific integral needed: \({\rm{I}}_1\).
• Substitute \({\rm{m}} = 1\) into the general integral definition.
• Use the trigonometric identity \(\sin 2{\rm{x}} = 2 \sin {\rm{x}} \cos {\rm{x}}\) to simplify the integrand.
• Cancel the common term \(\sin {\rm{x}}\).
• Evaluate the resulting simple definite integral \( \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos {\rm{x}} {\rm{dx}} \).
• The result of the evaluation is 0.

Connecting to Options

The calculated value for \({\rm{I}}_1\) is 0. Let's compare this with the given options:

• Option 1: 0
• Option 2: 1 / 2
• Option 3: 1
• Option 4: 2

Our result, 0, matches Option 1.

Revision Table: Integral Evaluation

Additional Information: Properties of I_m Integrals

The integral \({\rm{I}}_{\rm{m}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}}\) is an interesting type of integral. Let's explore some properties:

• For \({\rm{m}}=1\), we found \({\rm{I}}_1 = 0\).
• For \({\rm{m}}=2\), \({\rm{I}}_2 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 4{\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\). Using sum-to-product or multiple applications of double angle formulas, one can show that \(\frac{\sin 4x}{\sin x} = 4 \cos x \cos 2x\). Evaluating this integral is more complex but results in \({\rm{I}}_2 = \pi\).
• For \({\rm{m}}=3\), \({\rm{I}}_3 = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 6{\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\). Similarly, \(\frac{\sin 6x}{\sin x} = 2 \cos x + 2 \cos 3x + 2 \cos 5x\). Evaluating this results in \({\rm{I}}_3 = 0\).

It turns out there's a pattern for \({\rm{I}}_{\rm{m}}\):

• If m is an even positive integer (m = 2k), \({\rm{I}}_{\rm{m}} = \pi\).
• If m is an odd positive integer (m = 2k - 1), \({\rm{I}}_{\rm{m}} = 0\).

Since \({\rm{m}}=1\) is an odd integer, \({\rm{m}} = 2(1) - 1\), our result \({\rm{I}}_1 = 0\) aligns with this general property. This class of integrals is related to Dirichlet integrals and Fourier series.