1. We need to evaluate $I=\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\cos x}.$

2. Use the identity $1+\cos x=2\cos^2\frac{x}{2}.$ So, $\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}=\frac{1}{2}\sec^2\frac{x}{2}.$

Thus, $I=\int_{\pi/4}^{3\pi/4}\frac{1}{2}\sec^2\frac{x}{2}\,dx.$

3. Now use substitution: $u=\frac{x}{2} \quad\Rightarrow\quad dx=2\,du.$ Then

• when $x=\frac{\pi}{4}$, $u=\frac{\pi}{8}$,
• when $x=\frac{3\pi}{4}$, $u=\frac{3\pi}{8}$.

So, $I=\int_{\pi/8}^{3\pi/8}\frac{1}{2}\sec^2 u\,(2\,du)=\int_{\pi/8}^{3\pi/8}\sec^2 u\,du.$

4. Integrate: $\int \sec^2 u\,du=\tan u.$ Hence, $I=\tan\frac{3\pi}{8}-\tan\frac{\pi}{8}.$

5. Use standard values: $\tan\frac{\pi}{8}=\sqrt{2}-1,$ $\tan\frac{3\pi}{8}=\cot\frac{\pi}{8}=\frac{1}{\tan(\pi/8)}=\sqrt{2}+1.$

Therefore, $I=(\sqrt{2}+1)-(\sqrt{2}-1)=2.$

6. Compare with options:

• A: $2$ ✓
• B: $4$
• C: $-1$
• D: $-2$

So the correct option is A.