Question

If \( I_1 = \int_{-\pi/4}^{\pi/4} \frac{dx}{1 + \cos 2x} \) and \( I_2 = \int_{-1/2}^{1/2} |x| dx \), then show that \( I_1 - 4I_2 = 0 \).

Hint:

Checking for even/odd functions simplifies integrals over symmetric intervals significantly.
Always simplify the trigonometric denominator first before integrating.

Answer 1

Concept:
• Definite integral properties for even functions: \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
• Half-angle identity: \( 1 + \cos 2x = 2 \cos^2 x \).

Step 1: Evaluate \( I_1 \)
The function \( \frac{1}{1 + \cos 2x} \) is even. \[ I_1 = 2 \int_{0}^{\pi/4} \frac{dx}{2 \cos^2 x} = \int_{0}^{\pi/4} \sec^2 x \, dx \] \[ I_1 = \left[ \tan x \right]_{0}^{\pi/4} = \tan \frac{\pi}{4} - \tan 0 = 1 \]

Step 2: Evaluate \( I_2 \)
The function \( |x| \) is even. \[ I_2 = 2 \int_{0}^{1/2} x \, dx \] \[ I_2 = 2 \left[ \frac{x^2}{2} \right]_{0}^{1/2} = \left[ x^2 \right]_{0}^{1/2} \] \[ I_2 = (1/2)^2 - 0 = \frac{1}{4} \]

Step 3: Verify the given relationship
Calculate \( I_1 - 4I_2 \): \[ 1 - 4(1/4) = 1 - 1 = 0 \] Hence shown.