$\int\limits_{-\pi/4}^{\pi/4} \frac{dx}{1 + \cos 2x}$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
Let $I = \int\limits_{-\pi/4}^{\pi/4} \frac{dx}{1 + \cos 2x} = \int\limits_{-\pi/4}^{\pi/4} \frac{dx}{2 \cos^2 x} \quad [∵\cos 2x = 2\cos^2 x - 1]$
$= \frac{1}{2} \int\limits_{-\pi/4}^{\pi/4} \sec^2 x \, dx = \int\limits_{0}^{\pi/4} \sec^2 x \, dx = [\tan x]_0^{\pi/4} = 1$
$\left[ \int\limits_{-a}^{a} f(x) \, dx = 2 \int\limits_{0}^{a} f(x) \, dx \text{ if } f(-x) = f(x) \right]$