Question

The integral $ \int_{0}^{\pi} \frac{8x dx}{4 \cos^2 x + \sin^2 x} $ is equal to

• A. \( 2\pi^2 \)
• B. \( 4\pi^2 \)
• C. \( \pi^2 \)
• D. \( \frac{3\pi^2}{2} \)

Hint:

Use the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \) to simplify the integral. Then, divide the numerator and denominator by \( \cos^2 x \) to convert the integral into a form involving \( \tan x \) and \( \sec^2 x \), which can be solved using substitution. Remember to adjust the limits of integration accordingly.

Answer 1

The Correct Option is A

Let the integral be \( I \): \[ I = \int_{0}^{\pi} \frac{8x}{4 \cos^2 x + \sin^2 x} dx \] Using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \): \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 (\pi - x) + \sin^2 (\pi - x)} dx \] \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 (-\cos x)^2 + (\sin x)^2} dx \] \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 x + \sin^2 x} dx \] Adding the two expressions for \( I \): \[ 2I = \int_{0}^{\pi} \frac{8x + 8(\pi - x)}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = \int_{0}^{\pi} \frac{8x + 8\pi - 8x}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = \int_{0}^{\pi} \frac{8\pi}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = 8\pi \int_{0}^{\pi} \frac{1}{4 \cos^2 x + \sin^2 x} dx \] Divide numerator and denominator by \( \cos^2 x \): \[ 2I = 8\pi \int_{0}^{\pi} \frac{\sec^2 x}{4 + \tan^2 x} dx \] Since the integrand has a period of \( \pi \), we can write: \[ 2I = 8\pi \times 2 \int_{0}^{\pi/2} \frac{\sec^2 x}{4 + \tan^2 x} dx \] Let \( t = \tan x \), so \( dt = \sec^2 x dx \). When \( x = 0 \), \( t = 0 \). When \( x = \pi/2 \), \( t \rightarrow \infty \). \[ 2I = 16\pi \int_{0}^{\infty} \frac{dt}{4 + t^2} \] \[ 2I = 16\pi \times \frac{1}{2} \left[ \tan^{-1} \left( \frac{t}{2} \right) \right]_{0}^{\infty} \] \[ 2I = 8\pi \left( \tan^{-1} (\infty) - \tan^{-1} (0) \right) \] \[ 2I = 8\pi \left( \frac{\pi}{2} - 0 \right) \] \[ 2I = 4\pi^2 \] \[ I = 2\pi^2 \] The integral is equal to \( 2\pi^2 \).