Integrate(1+sinx)/(cosx)?

Question

the answer is ln !secx + tanX! + ln !secx! + c
HOW?

Answer 1

Seperate

INT sec x dx + INT tan x dx

ln | sec x + tan x | + ln | sec x | + K

(you just are supposed to know what those integrals are!)

Answer 2

∫ (1+sinx)/(cosx)dx divide out by cosx

= ∫ 1/cosx + ∫ sinx/cosx dx

= ∫ secx dx + ∫tanx dx

now ∫ secx = ln|secx + tanx|
because see http://math2.org/math/integrals/more/sec.htm

and ∫tanx dx = ln(secx)

Therefore answer = ln(sec x + tanx) + ln |sec x| + c

Answer 3

∫ (1+sinx)/(cosx)dx

= ∫ 1/cosx + tanx dx

= ∫ secx + tanx dx

= ln|sec + tanx| + ln |secx| + c

Answer 4

f'(x)=cosx-sinx f''(x)= -sinx-cosx by using fact the spinoff of sinx = cosx and the spinoff of cosx= -sinx via definition. set the equations =0 to locate your extreme factors, increasing/lowering, concave up/concave down.

Answer 5

I = ∫ sec x + tan x .dx
I = ∫sec x dx + ∫ tanx dx
I = Ia + Ib
I = log (sec x + tan x) - log cos x + C
Note Ia and Ib are listed in text books as standard integrals and may therefore be used without proof.