Consider the following for the next three (03) items that follow : Let \(I_1=\int_0^\pi \frac{x}{1+\cos ^2 x} d x\) and \(I_2=\int_0^\pi \frac{1}{1+\sin ^2 x} d x \)
What is the value of \(\frac{I_1+I_2}{I_1-I_2} \) ?
Understanding the Problem
The problem asks us to evaluate a specific ratio involving two definite integrals, \(I_1\) and \(I_2\). The integrals are defined over the interval \([0, \pi]\) and involve trigonometric functions.
- Integral \(I_1\): \( \int_0^\pi \frac{x}{1+\cos ^2 x} d x \)
- Integral \(I_2\): \( \int_0^\pi \frac{1}{1+\sin ^2 x} d x \)
We need to find the value of \( \frac{I_1+I_2}{I_1-I_2} \).
Evaluating Integral \(I_1\) using Integral Properties
We can use the property of definite integrals: \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). For \(I_1\), \(a=\pi\). Applying this property:
\( I_1 = \int_0^\pi \frac{\pi-x}{1+\cos ^2 (\pi-x)} d x \)
Since \( \cos(\pi-x) = -\cos x \), we have \( \cos^2(\pi-x) = (-\cos x)^2 = \cos^2 x \). So,
\( I_1 = \int_0^\pi \frac{\pi-x}{1+\cos ^2 x} d x \)
Now, add the original expression for \(I_1\) and this new expression:
\( I_1 + I_1 = \int_0^\pi \frac{x}{1+\cos ^2 x} d x + \int_0^\pi \frac{\pi-x}{1+\cos ^2 x} d x \)
\( 2I_1 = \int_0^\pi \frac{x + (\pi-x)}{1+\cos ^2 x} d x = \int_0^\pi \frac{\pi}{1+\cos ^2 x} d x \)
\( 2I_1 = \pi \int_0^\pi \frac{1}{1+\cos ^2 x} d x \)
Let \( J = \int_0^\pi \frac{1}{1+\cos ^2 x} d x \). So, \( 2I_1 = \pi J \).
Evaluating Integral \(I_2\)
Now let's evaluate \( I_2 = \int_0^\pi \frac{1}{1+\sin ^2 x} d x \). We can use the substitution \( t = \tan x \). However, \( \tan x \) is discontinuous at \( x = \pi/2 \). We split the integral at \( x = \pi/2 \).
\( I_2 = \int_0^{\pi/2} \frac{1}{1+\sin ^2 x} d x + \int_{\pi/2}^\pi \frac{1}{1+\sin ^2 x} d x \)
Consider the first part: \( \int_0^{\pi/2} \frac{1}{1+\sin ^2 x} d x \). Divide numerator and denominator by \( \cos^2 x \):
\( \int_0^{\pi/2} \frac{\sec^2 x}{\sec^2 x + \tan^2 x} d x = \int_0^{\pi/2} \frac{\sec^2 x}{(1+\tan^2 x) + \tan^2 x} d x = \int_0^{\pi/2} \frac{\sec^2 x}{1+2\tan^2 x} d x \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \). When \( x=0 \), \( t=\tan 0 = 0 \). When \( x \to \pi/2^- \), \( t \to \infty \).
\( \int_0^\infty \frac{dt}{1+2t^2} = \int_0^\infty \frac{dt}{1+(\sqrt{2}t)^2} \)
Let \( u = \sqrt{2}t \), so \( du = \sqrt{2} dt \). When \( t=0 \), \( u=0 \). When \( t \to \infty \), \( u \to \infty \).
\( \int_0^\infty \frac{du/\sqrt{2}}{1+u^2} = \frac{1}{\sqrt{2}} \int_0^\infty \frac{du}{1+u^2} = \frac{1}{\sqrt{2}} [\arctan u]_0^\infty = \frac{1}{\sqrt{2}} (\frac{\pi}{2} - 0) = \frac{\pi}{2\sqrt{2}} \)
Now consider the second part: \( \int_{\pi/2}^\pi \frac{1}{1+\sin ^2 x} d x \). Using the substitution \( t = \tan x \), when \( x \to \pi/2^+ \), \( t \to -\infty \). When \( x=\pi \), \( t=\tan \pi = 0 \).
\( \int_{-\infty}^0 \frac{dt}{1+2t^2} = [\frac{1}{\sqrt{2}} \arctan(\sqrt{2}t)]_{-\infty}^0 = \frac{1}{\sqrt{2}} (\arctan 0 - \lim_{v \to -\infty} \arctan v) = \frac{1}{\sqrt{2}} (0 - (-\frac{\pi}{2})) = \frac{\pi}{2\sqrt{2}} \)
So, \( I_2 = \frac{\pi}{2\sqrt{2}} + \frac{\pi}{2\sqrt{2}} = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}} \).
Evaluating Integral \(J = \int_0^\pi \frac{1}{1+\cos ^2 x} d x\)
Now let's evaluate \( J = \int_0^\pi \frac{1}{1+\cos ^2 x} d x \). Similar to \(I_2\), we split the integral and use \( t = \tan x \).
\( J = \int_0^{\pi/2} \frac{1}{1+\cos ^2 x} d x + \int_{\pi/2}^\pi \frac{1}{1+\cos ^2 x} d x \)
Consider the first part: \( \int_0^{\pi/2} \frac{1}{1+\cos ^2 x} d x \). Divide numerator and denominator by \( \cos^2 x \):
\( \int_0^{\pi/2} \frac{\sec^2 x}{\sec^2 x + 1} d x = \int_0^{\pi/2} \frac{\sec^2 x}{(1+\tan^2 x) + 1} d x = \int_0^{\pi/2} \frac{\sec^2 x}{2+\tan^2 x} d x \)
Let \( t = \tan x \), so \( dt = \sec^2 x dx \). When \( x=0 \), \( t=0 \). When \( x \to \pi/2^- \), \( t \to \infty \).
\( \int_0^\infty \frac{dt}{2+t^2} = \int_0^\infty \frac{dt}{(\sqrt{2})^2+t^2} = [\frac{1}{\sqrt{2}} \arctan(\frac{t}{\sqrt{2}})]_0^\infty = \frac{1}{\sqrt{2}} (\frac{\pi}{2} - 0) = \frac{\pi}{2\sqrt{2}} \)
Now consider the second part: \( \int_{\pi/2}^\pi \frac{1}{1+\cos ^2 x} d x \). Using the substitution \( t = \tan x \), when \( x \to \pi/2^+ \), \( t \to -\infty \). When \( x=\pi \), \( t=\tan \pi = 0 \).
\( \int_{-\infty}^0 \frac{dt}{2+t^2} = [\frac{1}{\sqrt{2}} \arctan(\frac{t}{\sqrt{2}})]_{-\infty}^0 = \frac{1}{\sqrt{2}} (\arctan 0 - \lim_{v \to -\infty} \arctan(\frac{v}{\sqrt{2}})) = \frac{1}{\sqrt{2}} (0 - (-\frac{\pi}{2})) = \frac{\pi}{2\sqrt{2}} \)
So, \( J = \frac{\pi}{2\sqrt{2}} + \frac{\pi}{2\sqrt{2}} = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}} \).
Relating \(I_1\), \(I_2\), and \(J\)
From the evaluation of \(I_2\) and \(J\), we found that \( I_2 = \frac{\pi}{\sqrt{2}} \) and \( J = \frac{\pi}{\sqrt{2}} \). Therefore, \( I_2 = J \).
From the property applied to \(I_1\), we had \( 2I_1 = \pi J \).
Substituting \( J = I_2 \) into the equation for \( 2I_1 \):
\( 2I_1 = \pi I_2 \)
This gives us a direct relationship between \(I_1\) and \(I_2\): \( I_1 = \frac{\pi}{2} I_2 \).
Calculating the Required Ratio
We need to find the value of \( \frac{I_1+I_2}{I_1-I_2} \). Substitute the relationship \( I_1 = \frac{\pi}{2} I_2 \) into this expression:
\( \frac{I_1+I_2}{I_1-I_2} = \frac{\frac{\pi}{2} I_2 + I_2}{\frac{\pi}{2} I_2 - I_2} \)
Assuming \( I_2 \neq 0 \) (which is true since the integrand is positive over the interval, so the integral is positive), we can factor out \( I_2 \) from the numerator and the denominator:
\( \frac{I_2 (\frac{\pi}{2} + 1)}{I_2 (\frac{\pi}{2} - 1)} = \frac{\frac{\pi}{2} + 1}{\frac{\pi}{2} - 1} \)
To simplify, find a common denominator in the numerator and denominator:
\( \frac{\frac{\pi+2}{2}}{\frac{\pi-2}{2}} \)
Multiply the numerator by the reciprocal of the denominator:
\( \frac{\pi+2}{2} \times \frac{2}{\pi-2} = \frac{\pi+2}{\pi-2} \)
Conclusion
The value of \( \frac{I_1+I_2}{I_1-I_2} \) is \( \frac{\pi+2}{\pi-2} \).
Summary of Integral Values (for reference)
From \( 2I_1 = \pi I_2 \) and \( I_2 = \frac{\pi}{\sqrt{2}} \), we can find \(I_1\):
\( 2I_1 = \pi \left(\frac{\pi}{\sqrt{2}}\right) = \frac{\pi^2}{\sqrt{2}} \)
\( I_1 = \frac{\pi^2}{2\sqrt{2}} \)
So, \( I_1 = \frac{\pi^2}{2\sqrt{2}} \) and \( I_2 = \frac{\pi}{\sqrt{2}} \).
| Integral | Value |
|---|---|
| \( I_1 \) | \( \frac{\pi^2}{2\sqrt{2}} \) |
| \( I_2 \) | \( \frac{\pi}{\sqrt{2}} \) |
Ratio Calculation Steps
- Start with the expression \( \frac{I_1+I_2}{I_1-I_2} \).
- Use the relationship \( I_1 = \frac{\pi}{2} I_2 \) derived from \( 2I_1 = \pi I_2 \).
- Substitute \( I_1 \) in the ratio expression: \( \frac{\frac{\pi}{2} I_2 + I_2}{\frac{\pi}{2} I_2 - I_2} \).
- Factor out \( I_2 \): \( \frac{I_2 (\frac{\pi}{2} + 1)}{I_2 (\frac{\pi}{2} - 1)} \).
- Cancel \( I_2 \) (since \( I_2 \neq 0 \)): \( \frac{\frac{\pi}{2} + 1}{\frac{\pi}{2} - 1} \).
- Simplify the complex fraction: \( \frac{\frac{\pi+2}{2}}{\frac{\pi-2}{2}} = \frac{\pi+2}{\pi-2} \).
Revision Table: Key Concepts
| Concept | Description | Application in Problem |
|---|---|---|
| Definite Integral Property | \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) | Used to relate \(I_1\) to an integral similar to \(I_2\). |
| Trigonometric Identities | \( \cos(\pi-x) = -\cos x \), \( \sec^2 x = 1+\tan^2 x \) | Used to simplify integrands after applying properties or substitutions. |
| Substitution Method | Replacing variable (e.g., \( t=\tan x \)) and adjusting integral limits and differential \( dt \). | Used to evaluate the standard integrals \( \int \frac{\sec^2 x}{a^2+\tan^2 x} dx \) and \( \int \frac{\sec^2 x}{a^2+b\tan^2 x} dx \). |
| Splitting Integrals | Breaking an integral over \([a, b]\) into \(\int_a^c + \int_c^b\) due to discontinuities or function definition changes. | Required because \( \tan x \) is discontinuous at \( \pi/2 \) in the interval \([0, \pi]\). |
Additional Information: Standard Integrals
The evaluation of \( \int_0^\pi \frac{1}{a+b\cos^2 x} dx \) and \( \int_0^\pi \frac{1}{a+b\sin^2 x} dx \) are common types in calculus. For integrals over \([0, \pi]\) or \([0, 2\pi]\), the substitution \( t=\tan x \) (with splitting if needed) or \( t=\tan(x/2) \) are frequently used techniques.
- Integral of the form \( \int_0^\pi \frac{1}{A+B\cos^2 x} dx \): Divide numerator and denominator by \( \cos^2 x \) to get \( \int_0^\pi \frac{\sec^2 x}{(A/ \cos^2 x) + B} dx = \int_0^\pi \frac{\sec^2 x}{A\sec^2 x + B} dx = \int_0^\pi \frac{\sec^2 x}{A(1+\tan^2 x) + B} dx \). Then use \( t=\tan x \) and split the interval \([0, \pi]\) into \([0, \pi/2]\) and \([\pi/2, \pi]\).
- Integral of the form \( \int_0^\pi \frac{1}{A+B\sin^2 x} dx \): Similarly, divide by \( \cos^2 x \) to get \( \int_0^\pi \frac{\sec^2 x}{(A/ \cos^2 x) + (B\sin^2 x / \cos^2 x)} dx = \int_0^\pi \frac{\sec^2 x}{A\sec^2 x + B\tan^2 x} dx = \int_0^\pi \frac{\sec^2 x}{A(1+\tan^2 x) + B\tan^2 x} dx = \int_0^\pi \frac{\sec^2 x}{A+(A+B)\tan^2 x} dx \). Then use \( t=\tan x \) and split the interval.
In this problem, \( J = \int_0^\pi \frac{1}{1+\cos^2 x} dx \) corresponds to \( A=1, B=1 \), and \( I_2 = \int_0^\pi \frac{1}{1+\sin^2 x} dx \) corresponds to \( A=1, B=1 \). Both evaluate to \( \frac{\pi}{\sqrt{2}} \).
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