Question

If \( I = \int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \), find the value of I.

Hint:

For definite integrals of the form \( \int_{0}^{a} x f(x) \, dx \), always try applying the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \) first. It often simplifies the problem by eliminating the \( x \) term in the numerator.

Answer 1

Step 1: Understanding the Concept:
This is a standard form of a definite integral that can be solved using a property of definite integrals, often known as the "King's Rule", which helps to eliminate the 'x' in the numerator.
Step 2: Key Formula or Approach:
We use the property: \( \int_{0}^{c} f(x) \, dx = \int_{0}^{c} f(c-x) \, dx \).
Let the given integral be: \[ I = \int_{0}^{\pi} \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \quad \cdots (1) \] Applying the property with \( c = \pi \): \[ I = \int_{0}^{\pi} \frac{\pi - x}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)} \, dx \] Since \( \cos(\pi-x) = -\cos x \) and \( \sin(\pi-x) = \sin x \), we have \( \cos^2(\pi-x) = \cos^2 x \) and \( \sin^2(\pi-x) = \sin^2 x \). \[ I = \int_{0}^{\pi} \frac{\pi - x}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \quad \cdots (2) \] Step 3: Detailed Explanation:
Add equations (1) and (2): \[ 2I = \int_{0}^{\pi} \frac{x + (\pi - x)}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx = \int_{0}^{\pi} \frac{\pi}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] \[ 2I = \pi \int_{0}^{\pi} \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] Let the integrand be \( g(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \). Since \( g(\pi-x) = g(x) \), we can use the property \( \int_{0}^{2c} f(x) dx = 2 \int_{0}^{c} f(x) dx \). Here \( 2c = \pi \Rightarrow c = \pi/2 \). \[ 2I = \pi \left( 2 \int_{0}^{\pi/2} \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \right) = 2\pi \int_{0}^{\pi/2} \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] \[ I = \pi \int_{0}^{\pi/2} \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx \] Divide the numerator and denominator by \( \cos^2 x \): \[ I = \pi \int_{0}^{\pi/2} \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} \, dx \] Let \( t = \tan x \), so \( dt = \sec^2 x \, dx \). The limits of integration 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 of the form \( \int \frac{1}{c^2+u^2}du = \frac{1}{c} \tan^{-1}(\frac{u}{c}) \). \[ I = \pi \left[ \frac{1}{a} \tan^{-1}\left(\frac{bt}{a}\right) \cdot \frac{1}{b} \right]_{0}^{\infty} = \frac{\pi}{ab} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_{0}^{\infty} \] \[ I = \frac{\pi}{ab} \left( \lim_{t \to \infty} \tan^{-1}\left(\frac{bt}{a}\right) - \tan^{-1}(0) \right) \] \[ I = \frac{\pi}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{2ab} \] Step 4: Final Answer:
The value of the integral is \( I = \frac{\pi^2}{2ab} \).