Question

Evaluate: \( \int_0^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \)

Hint:

Definite integrals of the form \( \int_0^a x f(x) dx \) can often be simplified using the "King's property": \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). This is a very powerful tool for eliminating the 'x' term.

Answer 1

Step 1: Understanding the Concept:
This definite integral can be solved using the properties of definite integrals. The presence of 'x' in the numerator suggests using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), which often helps to eliminate the 'x' term.
Step 2: Key Formula or Approach:
1. Use the property \( I = \int_0^{\pi} f(x) dx = \int_0^{\pi} f(\pi - x) dx \).
2. Add the two forms of the integral to eliminate x from the numerator.
3. Use the property \( \int_0^{2a} g(x) dx = 2 \int_0^a g(x) dx \) if \( g(2a-x) = g(x) \).
4. Evaluate the resulting integral by dividing the numerator and denominator by \( \cos^2 x \) and using a substitution.
Step 3: Detailed Explanation or Calculation:
Let the given integral be \( I \). \[ I = \int_0^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} . (1) \] Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \):
\[ I = \int_0^{\pi} \frac{(\pi - x) \, dx}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)} \] Since \( \cos(\pi-x) = -\cos x \) and \( \sin(\pi-x) = \sin x \), their squares are \( \cos^2 x \) and \( \sin^2 x \). \[ I = \int_0^{\pi} \frac{(\pi - x) \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} . (2) \] Adding equations (1) and (2):
\[ 2I = \int_0^{\pi} \frac{x + (\pi - x)}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx = \int_0^{\pi} \frac{\pi}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] \[ I = \frac{\pi}{2} \int_0^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \] Let \( g(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \). Since \( g(\pi - x) = g(x) \), we can use the property \( \int_0^{2a} g(x) dx = 2 \int_0^a g(x) dx \). Here \( 2a = \pi \). \[ I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \pi \int_0^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \] Divide numerator and denominator by \( \cos^2 x \): \[ I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{a^2 + b^2 \tan^2 x} \] Let \( t = \tan x \), so \( dt = \sec^2 x \, dx \). The limits change from \( x=0 \to t=0 \) and \( x=\pi/2 \to t=\infty \). \[ I = \pi \int_0^{\infty} \frac{dt}{a^2 + (bt)^2} \] This is a standard integral form: \[ I = \pi \left[ \frac{1}{b} \cdot \frac{1}{a} \tan^{-1}\left(\frac{bt}{a}\right) \right]_0^{\infty} = \frac{\pi}{ab} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_0^{\infty} \] \[ I = \frac{\pi}{ab} \left( \lim_{t \to \infty} \tan^{-1}\left(\frac{bt}{a}\right) - \tan^{-1}(0) \right) = \frac{\pi}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{2ab} \] Step 4: Final Answer:
The value of the integral is \( \frac{\pi^2}{2ab} \).