Question

${\int }_{-\pi /4}^{\pi /4}\frac{dx}{1+\cos 2x}$ is equal to

• A. $2$
• B. $1$
• C. $4$
• D. $0$

Answer 1

Answer: (B)

$1$

Let $I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1+\cos 2x}$

$={\int }_{-\pi /4}^{\pi /4}\frac{dx}{2{\cos }^{2}x}$

$=\frac{1}{2}\cdot 2{\int }_0^{\pi /4}\frac{dx}{{\cos }^{2}x}$, since $\frac{1}{{\cos }^{2}x}$ is an even function

$={\int }_0^{\pi /4}{\sec }^{2}xdx=[\tan x{]}_0^{\pi /4}=\tan \frac{\pi }{4}-\tan (0)=1-0=1$