integrate:cosx+2/cosx+sinx?????

Answer 1

As written, you have

∫cosx + 2/cosx + sinx dx
which is just trivial:
sinx + 2ln(secx+tanx) - cosx + C

So, I'll assume you meant

∫(cosx + 2)/(cosx + sinx) dx
That's tougher than it looks.

cosx/(cosx+sinx)
= 2cosx/2(cosx+sinx)
= (cosx+sinx + cosx-sinx)/2(cosx+sinx)
= (cosx+sinx)/2(cosx+sinx) + (cosx-sinx)/2(cosx+sinx)
= 1/2 + 1/2 (cosx-sinx)/(cosx+sinx)

Now take take the integral, and let
u = cosx+sinx
du = cosx-sinx dx
and you have

∫ 1/2 dx + ∫du/2u
= 1/2 x + 1/2 log(u)
= 1/2 (x + log(cosx+sinx)) + C

Now see what you can do with the other part. Recall that

cosx+sinx = √2 sin(x+π/4)