Analyzing the Given Integral and Statements

The question asks us to consider a definite integral 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 then presented with two statements about this integral and asked to determine which one is correct:

1. \( {{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}}\) is equal to 0
2. \( {{\rm{I}}_{2{\rm{m}}}} > {\rm{\;}}{{\rm{I}}_{\rm{m}}}\)

Evaluating Statement 1: \({\rm{I}}_{\rm{m}} - {\rm{I}}_{{\rm{m}} - 1} = 0\)

Let's evaluate the difference between \(I_m\) and \(I_{m-1}\):

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}} - \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2\left( {{\rm{m}} - 1} \right){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\)

We can combine these two integrals since they have the same limits and denominator:

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}} - \sin 2\left( {{\rm{m}} - 1} \right){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\)

Now, we use the trigonometric identity for the difference of sines: \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \).

Let \(A = 2mx\) and \(B = 2(m-1)x\). Then:

\(\frac{A+B}{2} = \frac{2mx + 2(m-1)x}{2} = \frac{2mx + 2mx - 2x}{2} = \frac{4mx - 2x}{2} = (2m - 1)x\)

\(\frac{A-B}{2} = \frac{2mx - 2(m-1)x}{2} = \frac{2mx - 2mx + 2x}{2} = \frac{2x}{2} = x\)

So, \( \sin 2mx - \sin 2(m-1)x = 2 \cos((2m-1)x) \sin x \).

Substituting this back into the integral:

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{2 \cos \left( {\left( {2{\rm{m}} - 1} \right){\rm{x}}} \right)\sin {\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}}\)

For \(x \in (0, \pi)\), \(\sin x \neq 0\), so we can cancel the \(\sin x\) terms:

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos \left( {\left( {2{\rm{m}} - 1} \right){\rm{x}}} \right){\rm{dx}}\)

Now, we evaluate this integral:

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = 2 \left[ \frac{{\sin \left( {\left( {2{\rm{m}} - 1} \right){\rm{x}}} \right)}}{{2{\rm{m}} - 1}} \right]_0^{\rm{\pi }}\)

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = 2 \left( \frac{{\sin \left( {\left( {2{\rm{m}} - 1} \right){\rm{\pi }}} \right)}}{{2{\rm{m}} - 1}} - \frac{{\sin \left( {\left( {2{\rm{m}} - 1} \right)0} \right)}}{{2{\rm{m}} - 1}} \right)\)

Since \(m\) is a positive integer, \(2m-1\) is always an odd integer (\(1, 3, 5, \dots\)). The sine of an odd multiple of \(\pi\) is 0, i.e., \(\sin((2m-1)\pi) = 0\). Also, \(\sin(0) = 0\).

\({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = 2 \left( \frac{0}{{2{\rm{m}} - 1}} - \frac{0}{{2{\rm{m}} - 1}} \right) = 2(0 - 0) = 0\)

Thus, statement 1 is correct: \( {{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = 0 \).

Evaluating Statement 2: \({\rm{I}}_{2{\rm{m}}} > {\rm{I}}_{\rm{m}}\)

From the evaluation of statement 1, we found that \(I_m - I_{m-1} = 0\). This implies \(I_m = I_{m-1}\) for any positive integer \(m \ge 2\). Since this relation holds for consecutive integer values of \(m\), it means that \(I_m\) is constant for all positive integers \(m\).

Let's calculate the value of this constant by finding \(I_1\):

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

Using the identity \(\sin 2x = 2 \sin x \cos x\):

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

For \(x \in (0, \pi)\), \(\sin x \neq 0\), so we can cancel \(\sin x\):

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

Now, we evaluate the integral:

\({{\rm{I}}_1} = \left[ {2 \sin {\rm{x}}} \right]_0^{\rm{\pi }} = 2 \sin {\rm{\pi }} - 2 \sin 0 = 2(0) - 2(0) = 0\)

Since \(I_m = I_{m-1}\) for \(m \ge 2\), and \(I_1 = 0\), we can conclude that \(I_m = 0\) for all positive integers \(m\).

Now let's check statement 2: \( {{\rm{I}}_{2{\rm{m}}}} > {\rm{\;}}{{\rm{I}}_{\rm{m}}} \).

Since \(I_m = 0\) for all positive integers \(m\), we have \(I_{2m} = 0\) (as \(2m\) is also a positive integer if \(m\) is) and \(I_m = 0\).

The statement becomes \(0 > 0\), which is false.

Thus, statement 2 is incorrect.

Conclusion

Based on our analysis:

• Statement 1: \( {{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}} = 0 \) is correct.
• Statement 2: \( {{\rm{I}}_{2{\rm{m}}}} > {\rm{\;}}{{\rm{I}}_{\rm{m}}} \) is incorrect.

Therefore, only statement 1 is correct.

Revision Table: Definite Integral Properties

Additional Information: Symmetry Properties of Integrals

The integral \(\mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}}\) can also be analyzed using symmetry properties. For example, consider the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\).

Let \(f(x) = \frac{\sin 2mx}{\sin x}\). Then \(f(\pi - x) = \frac{\sin(2m(\pi - x))}{\sin(\pi - x)} = \frac{\sin(2m\pi - 2mx)}{\sin x}\).

Using \(\sin(n\pi - \theta) = (-1)^{n+1}\sin \theta\), for integer \(n=2m\):

\(\sin(2m\pi - 2mx) = (-1)^{2m+1} \sin(2mx) = (-1)^{odd} \sin(2mx) = -\sin(2mx)\).

So, \(f(\pi - x) = \frac{-\sin 2mx}{\sin x} = -f(x)\).

This means the integrand is odd about the center of the interval \(x = \pi/2\). However, the integral is from 0 to \(\pi\), which is symmetric about \(\pi/2\).

For an integral \(\int_0^{2a} f(x) dx\), if \(f(2a-x) = -f(x)\), then \(\int_0^{2a} f(x) dx = 0\).

In our case, \(2a = \pi\), so \(a = \pi/2\). We found \(f(\pi - x) = -f(x)\). Thus, \(\int_0^\pi \frac{\sin 2mx}{\sin x} dx = 0\). This confirms our finding that \(I_m = 0\) for all positive integers \(m\).