Question

hello, can someone please explain to me how the integral of sec^2(x) / tan(x) dx, = ln tan(x) + c ?

I don't understand how to simplify sec^2(x)/tan(x) to something that integrates into that.

thanks very much

Answer 1

u= tanx
du=sec^2 x dx

so integral of du/u= lnu
since u is tanx
so ln tanx

Answer 2

ln tanx +c ofcourse...i forgot that

Answer 3

Recall that:
\[\cos^2 x + sin^2 x = 1\]Dividing by \(cos^2 x\)\[ \implies1 + tan^2x = sec^2 x\]

\[\implies \int {sec^2x \over tan x} dx = \int {1+tan^2x \over tan x}dx\]

Answer 4

\[ = \int [\frac{1}{tanx} + tanx] dx\]

Which you can integrate term by term.

Answer 5

Or you can use a u-sub. That way is probably faster

Answer 6

thank you both.