Question

Integrate?

`(secx+tanx)dx`

Answer 1

`int(secx+tanx)dx=lnabs(secx+tanx)-lnabscosx+C`

Explanation:

.

`int(secx+tanx)dx=intsecxdx+inttanxdx`

`intsecxdx=intsecx((secx+tanx)/(secx+tanx))dx=int(sec^2x+secxtanx)/(secx+tanx)dx`

Let `u=secx+tanx`

`du=(secxtanx+sec^2x)dx`

Now, we substitute the `u` and `du`:

`intsecxdx=int(du)/u=lnabsu+C`

Now we substitute back for `u`:

`intsecxdx=lnabs(secx+tanx)+C`

`inttanxdx=intsinx/cosxdx`

Let `u=cosx`

`du=-sinxdx`

Now, we substitute `u` and `du`:

`inttanxdx=int-(du)/u=-int(du)/u=-lnabsu+C`

Now, we substitute back for `u`:

`inttanxdx=-lnabscosx+C`

Therefore,

`int(secx+tanx)dx=lnabs(secx+tanx)-lnabscosx+C`