The integral $$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x}$$ is equal to

We have to evaluate the definite integral

$$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos x}\;.$$

First, we recall the half-angle identity for cosine:

$$1+\cos x = 2\cos^2\frac{x}{2}\,.$$

Substituting this expression in the denominator, the integrand becomes

$$\frac{1}{1+\cos x} \;=\; \frac{1}{2\cos^2\frac{x}{2}} \;=\; \frac{1}{2}\sec^2\frac{x}{2}\,.$$

Hence the integral can be rewritten as

$$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos x} \;=\; \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{2}\sec^2\frac{x}{2}\,dx\,.$$

We bring the constant $$\dfrac12$$ outside:

$$=\; \frac12 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2\frac{x}{2}\,dx\,.$$

Now we notice that the derivative of $$\tan\frac{x}{2}$$ is $$\dfrac12\sec^2\frac{x}{2}$$, i.e.

$$\frac{d}{dx}\left(\tan\frac{x}{2}\right) = \frac12\sec^2\frac{x}{2}\,.$$

Therefore

$$\sec^2\frac{x}{2}\,dx = 2\,d\!\left(\tan\frac{x}{2}\right).$$

Substituting this differential in the integral gives

$$\frac12 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2\frac{x}{2}\,dx \;=\; \frac12 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2\, d\!\left(\tan\frac{x}{2}\right) \;=\; \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} d\!\left(\tan\frac{x}{2}\right)\,.$$

This integral of a total differential is simply the difference of the antiderivative at the limits:

$$= \tan\frac{x}{2}\Bigg|_{x=\frac{\pi}{4}}^{x=\frac{3\pi}{4}} \;=\; \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right).$$

To proceed, we evaluate the two tangent values. A well-known exact result for the half-angle of $$45^\circ$$ states

$$\tan\frac{\pi}{8} = \sqrt2 - 1\;,\qquad \tan\frac{3\pi}{8} = \sqrt2 + 1.$$

Substituting these values, we get

$$\tan\frac{3\pi}{8} - \tan\frac{\pi}{8} = \left(\sqrt2 + 1\right) - \left(\sqrt2 - 1\right) = \sqrt2 + 1 - \sqrt2 + 1 = 2.$$

Thus the definite integral evaluates to

$$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = 2.$$

Hence, the correct answer is Option B.