## Calculus (3rd Edition)

We have $$\int_{0}^{\pi/2} \tan x dx=\int_{0}^{\pi/2} \frac{\sin x}{\cos x} dx\\ =-\int_{1}^{0} \frac{d( \cos x)}{\cos x} =\int_{0}^{1} \frac{d( \cos x)}{\cos x}\\ =\ln(\cos x)|_0^{\pi/2}=\ln 1- \ln 0=-\infty.$$ Hence, the integral diverges.