Given that \( I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x} \)
We know that, \( 1+\cos 2 x=2 \cos ^{2} x \)
So, \( I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{2 \cos ^{2} x}=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \)
We know that if \( \mathrm{f}(\mathrm{x}) \) is even function then,
\[
\begin{array}{l}
\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\
\text { So, } I=\frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x=\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\
=[\tan x]_{0}^{\frac{r}{4}}=\tan \frac{\Pi}{4}-\tan 0=1
\end{array}
\]