Direction: Consider the integral: \({{\rm{I}}_{\rm{m}}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}},\) where m is a positive integer.
What is I 2+ I 3equal to?
0
Understanding the Definite Integral
The problem asks us to evaluate the sum of two definite integrals, \(I_2\) and \(I_3\), where the integral \(I_m\) is defined as:
\[{{\rm{I}}_{\rm{m}}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}}\]for a positive integer \(m\).
Analyzing the Integral \(I_m\)
Let's investigate the relationship between \(I_{m+1}\) and \(I_m\). Consider the difference \(I_{m+1} - I_m\):
\[I_{m+1} - I_m = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2\left( {m + 1} \right){\rm{x}}}}{{\sin {\rm{x}}}}{\rm{dx}} - \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin 2{\rm{mx}}}}{{\sin {\rm{x}}}}{\rm{dx}}\] \[I_{m+1} - I_m = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin \left( {2\left( {m + 1} \right){\rm{x}}} \right) - \sin \left( {2{\rm{mx}}} \right)}}{{{\sin {\rm{x}}}}}{\rm{dx}}\]We can 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 \(A = 2(m+1)x\) and \(B = 2mx\). Then:
- \(\frac{A+B}{2} = \frac{2(m+1)x + 2mx}{2} = \frac{2mx + 2x + 2mx}{2} = \frac{4mx + 2x}{2} = (2m+1)x\)
- \(\frac{A-B}{2} = \frac{2(m+1)x - 2mx}{2} = \frac{2mx + 2x - 2mx}{2} = \frac{2x}{2} = x\)
So, \(\sin(2(m+1)x) - \sin(2mx) = 2 \cos((2m+1)x) \sin x\). Substituting this back into the integral difference:
\[I_{m+1} - I_m = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{2 \cos \left( {\left( {2m + 1} \right){\rm{x}}} \right)\sin {\rm{x}}}}{{{\sin {\rm{x}}}}}{\rm{dx}}\]For \(x \in (0, \pi)\), \(\sin x \neq 0\), so we can cancel \(\sin x\):
\[I_{m+1} - I_m = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos \left( {\left( {2m + 1} \right){\rm{x}}} \right){\rm{dx}}\]Now, we evaluate this integral:
\[I_{m+1} - I_m = 2 \left[ {\frac{{\sin \left( {\left( {2m + 1} \right){\rm{x}}} \right)}}{{2m + 1}}} \right]_0^\pi\] \[I_{m+1} - I_m = 2 \left( {\frac{{\sin \left( {\left( {2m + 1} \right)\pi } \right)}}{{2m + 1}} - \frac{{\sin \left( {\left( {2m + 1} \right) \cdot 0} \right)}}{{2m + 1}}} \right)\]Since \(m\) is a positive integer, \(2m+1\) is an integer. The sine of any integer multiple of \(\pi\) is \(0\). Thus, \(\sin((2m+1)\pi) = 0\) and \(\sin(0) = 0\).
\[I_{m+1} - I_m = 2 \left( {\frac{0}{{2m + 1}} - \frac{0}{{2m + 1}}} \right) = 0\]This result shows that \(I_{m+1} = I_m\) for all positive integers \(m\). This means the value of \(I_m\) is constant for all positive integers \(m\).
Calculating the Value of \(I_m\)
Since \(I_m\) is constant, we can find its value by calculating \(I_1\).
\[{I_1} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin \left( {2 \cdot 1 \cdot {\rm{x}}} \right)}}{{{\sin {\rm{x}}}}}{\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{\sin \left( {2{\rm{x}}} \right)}}{{{\sin {\rm{x}}}}}{\rm{dx}}\]Using the double angle formula, \(\sin(2x) = 2 \sin x \cos x\):
\[{I_1} = \mathop \smallint \nolimits_0^{\rm{\pi }} \frac{{2 \sin {\rm{x}} \cos {\rm{x}}}}{{{\sin {\rm{x}}}}}{\rm{dx}}\]For \(x \in (0, \pi)\), \(\sin x \neq 0\), so we can cancel \(\sin x\):
\[{I_1} = \mathop \smallint \nolimits_0^{\rm{\pi }} 2 \cos {\rm{x}} {\rm{dx}}\]Now, we evaluate this integral:
\[{I_1} = 2 \left[ {\sin {\rm{x}}} \right]_0^\pi\] \[{I_1} = 2 \left( {\sin \pi - \sin 0} \right)\] \[{I_1} = 2 \left( {0 - 0} \right) = 0\]Since \(I_1 = 0\) and \(I_m\) is constant for all positive integers \(m\), it follows that \(I_m = 0\) for all positive integers \(m\). This includes \(I_2\) and \(I_3\).
- \(I_2 = 0\)
- \(I_3 = 0\)
Calculating \(I_2 + I_3\)
Finally, we need to find the value of \(I_2 + I_3\):
\[{I_2} + {I_3} = 0 + 0 = 0\]The sum \(I_2 + I_3\) is equal to 0.
Summary of Results
| Integral | Value |
|---|---|
| \(I_m\) for any positive integer m | \(0\) |
| \(I_1\) | \(0\) |
| \(I_2\) | \(0\) |
| \(I_3\) | \(0\) |
| \(I_2 + I_3\) | \(0\) |
Conclusion
Based on our analysis, the value of \(I_m\) is \(0\) for any positive integer \(m\). Therefore, \(I_2 = 0\) and \(I_3 = 0\), and their sum \(I_2 + I_3\) is also \(0\).
Revision Table: Definite Integral Properties
| Concept | Description | Relevance to Problem |
|---|---|---|
| Definite Integral | Integral evaluated between upper and lower limits. Represents signed area under the curve. | The problem involves evaluating a specific definite integral \(I_m\). |
| Trigonometric Identities | Equations involving trigonometric functions (e.g., sum-to-product, double angle). | Used to simplify the integrand \(\frac{\sin 2mx}{\sin x}\) and the difference \(I_{m+1}-I_m\). |
| Recursive Relation | A formula that defines the terms of a sequence using previous terms. | We found a recursive relation \(I_{m+1} - I_m = 0\), showing the integral is constant. |
Additional Information: Generalizing the Integral
The integral \(\int_0^\pi \frac{\sin(nx)}{\sin x} dx\) is a known result. For a positive integer \(n\), its value depends on whether \(n\) is odd or even.
- If \(n\) is odd, \(\int_0^\pi \frac{\sin(nx)}{\sin x} dx = \pi\).
- If \(n\) is even, \(\int_0^\pi \frac{\sin(nx)}{\sin x} dx = 0\).
In our problem, the integral is \(I_m = \int_0^\pi \frac{\sin(2mx)}{\sin x} dx\). Here, the coefficient of \(x\) in the numerator is \(2m\). Since \(m\) is a positive integer, \(2m\) is always an even positive integer (specifically, \(2, 4, 6, \dots\)).
According to the general result for even \(n\), \(\int_0^\pi \frac{\sin(2mx)}{\sin x} dx = 0\). This confirms our finding that \(I_m = 0\) for all positive integers \(m\).
This general result provides a quicker way to see that \(I_2 = 0\) (since \(2m=4\), which is even) and \(I_3 = 0\) (since \(2m=6\), which is even), leading directly to \(I_2 + I_3 = 0 + 0 = 0\). The method using the difference \(I_{m+1}-I_m\) is a way to derive this specific case of the general result.
Important Questions from Definite Integrals
What is \(\displaystyle \int_0^\pi\left(\sin ^4 x+\cos ^4 x\right) d x\) equal to?
What is I equal to?
What is I 1equal to?
What is I m is equal to?
Consider the following:
1. \({{\rm{I}}_{\rm{m}}} - {{\rm{I}}_{{\rm{m}} - 1}}\) is equal to 0
2. \({{\rm{I}}_{2{\rm{m}}}} > {\rm{\;}}{{\rm{I}}_{\rm{m}}}\)
Which of the above is/are correct?
