Question

what is the integral of tanIx)

Answer 1

is it tanlx or tan ln(x)

Answer 2

-ln(cos[x])+C
\[ \int\text {tanx dx} \Rightarrow \int\text{sin[x]/cos[x] dx}\] Use a U-substitution of u cos[x] and du = -sin[x]dx and du/-sin[x] = dx so we have
\[\int\text{sin[x]/u (du/(-sin[x]))} \Rightarrow -\int\text{du/u => -ln[u]+C => -ln(cos[[x])}+C\]