12) \( \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) d x \)

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) d x \), we can use the substitution \( u = \frac{x}{2} \), which implies \( d x = 2 du \). The limits of integration change as follows: when \( x = 0 \), \( u = 0 \); and when \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \). Rewriting the integral, we have: \[ \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) d x = 2 \int_{0}^{\frac{\pi}{4}} \tan(u) du. \] Next, we need to evaluate \( \int \tan(u) du \). The integral of \(\tan(u)\) is given by: \[ \int \tan(u) du = -\ln|\cos(u)| + C. \] Thus, \[ \int_{0}^{\frac{\pi}{4}} \tan(u) du = \left[-\ln|\cos(u)|\right]_{0}^{\frac{\pi}{4}}. \] Evaluating this expression at the bounds, we find: \[ -\ln|\cos(\frac{\pi}{4})| - (-\ln|\cos(0)|) = -\ln\left(\frac{1}{\sqrt{2}}\right) - (-\ln(1)) = -\ln\left(\frac{1}{\sqrt{2}}\right) = \ln(\sqrt{2}). \] Therefore, \[ \int_{0}^{\frac{\pi}{4}} \tan(u) du = \ln(\sqrt{2}). \] Now, substituting back into our expression gives us: \[ 2 \int_{0}^{\frac{\pi}{4}} \tan(u) du = 2 \ln(\sqrt{2}) = \ln(2). \] In conclusion, the value of the original integral is: \[ \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) d x = \ln(2). \]