$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x}$ is equal to |
1 2 3 4 |
1 |
The correct answer is Option (1) - 1 $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x}=\frac{1}{2}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2xdx$ $=\frac{1}{2}\left|\tan x\right|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=1$ |