29. \( \int_{0}^{\pi} \tan x d x= \) A \( \frac{1}{2} \ln 2 \) B \( -\frac{1}{2} \ln 2 \) C \( \frac{1}{2} \ln 2-1 \) \( D-\frac{1}{2} \ln 2-1 \)

To evaluate the integral \( \int_{0}^{\pi} \tan x \, dx \), we can use the property of the tangent function and the symmetry of the integral. First, we note that: \[ \tan(\pi - x) = -\tan x \] This means that the integral can be split as follows: \[ \int_{0}^{\pi} \tan x \, dx = \int_{0}^{\frac{\pi}{2}} \tan x \, dx + \int_{\frac{\pi}{2}}^{\pi} \tan x \, dx \] For the second integral, we can make a substitution \( u = \pi - x \), which gives \( du = -dx \). The limits change as follows: when \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{2} \) and when \( x = \pi \), \( u = 0 \). Thus, we have: \[ \int_{\frac{\pi}{2}}^{\pi} \tan x \, dx = \int_{\frac{\pi}{2}}^{0} \tan(\pi - u) (-du) = \int_{0}^{\frac{\pi}{2}} -\tan u \, du = -\int_{0}^{\frac{\pi}{2}} \tan u \, du \] Combining these results, we find: \[ \int_{0}^{\pi} \tan x \, dx = \int_{0}^{\frac{\pi}{2}} \tan x \, dx - \int_{0}^{\frac{\pi}{2}} \tan x \, dx = 0 \] However, we need to be careful because \( \tan x \) has a vertical asymptote at \( x = \frac{\pi}{2} \). Therefore, we should evaluate the integral as an improper integral: \[ \int_{0}^{\pi} \tan x \, dx = \lim_{a \to \frac{\pi}{2}^-} \left( \int_{0}^{a} \tan x \, dx + \int_{a}^{\pi} \tan x \, dx \right) \] The integral \( \int_{0}^{a} \tan x \, dx \) can be computed as follows: \[ \int \tan x \, dx = -\ln |\cos x| + C \] Thus, \[ \int_{0}^{a} \tan x \, dx = -\ln |\cos a| + \ln |\cos 0| = -\ln |\cos a| + 0 = -\ln |\cos a| \] As \( a \to \frac{\pi}{2}^- \), \( \cos a \to 0 \) and \( -\ln |\cos a| \to \infty \). Therefore, the integral diverges to \( +\infty \). For the integral from \( a \) to \( \pi \): \[ \int_{a}^{\pi} \tan x \, dx = \lim_{b \to \pi^-} \left( -\ln |\cos x| \bigg|_{a}^{b} \right) = -\ln |\cos b| + \ln |\cos a| \] As \( b \to \pi \), \( \cos b \to -1 \) and \( -\ln |\cos b| \to \ln 1 = 0 \). Thus, we have: \[ \int_{a}^{\pi} \tan x \, dx = 0 + \ln |\cos a| \to -\infty \text{ as } a \to \frac{\pi}{2}^- \] Combining these results, we find that: \[ \int_{0}^{\pi} \tan x \, dx = +\infty - \infty \] This indicates that the integral does not converge to a finite value. Therefore, the integral \( \int_{0}^{\pi} \tan x \, dx \) diverges. Since the integral diverges, none of the provided options (A, B, C, D) are correct. The integral does not yield a finite result.