Let I = ∫tanx dx

=> I = ∫(sinx/cosx) dx

Let u = cosx

=> du =  -sinx dx

=> -du = sinx dx

Now, I = -∫du/u 

=> I = -log|u| + C

=> I = -log|cosx| + C

So, ∫tanx dx = -log|cosx| + C