## eminem 5 years ago what is the integral of tanIx)

1. anonymous

is it tanlx or tan ln(x)

2. anonymous

-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 $\int\text{sin[x]/u (du/(-sin[x]))} \Rightarrow -\int\text{du/u => -ln[u]+C => -ln(cos[[x])}+C$