how to solve for the integral of 1 over the cos(x) + 1?

Use the identity (cosx+1)/2=cos^2(x/2).

Then S=Sdx/[2cos^2(x/2)]=S(1/cos^2(x/2)) d(x/2)=tan(x/2)

The simplest you can get!!!

∫(1/cos x )+ 1dx = ∫ sec x dx + x + C

= log(sec x + tan x) + x + C

Note that the integral of sec x is a standard integral, log(sec x + tan x)

I would multiply 1/(cosx + 1) by (cosx -1)/(cosx - 1).

Then you will have

(cosx - 1)/(cos²x - 1),

(cosx - 1)/(-sin²x),

(-cosx)/(sin²x) + (1)/(sin²x),

-cotxcscx + csc²x.

The derivative of cscx is -cotxcscx and the derivative of cotx is csc²x.

Feel free to email me for further clarifications.

I assume you mean 1/(cos(x) + 1).

First let's rearrange the terms to make it easier to integrate.

1/(cos(x) + 1) = 1/[1 + cos(x)]

= [1 - cos(x)]/{[1 + cos(x)][1 - cos(x)]}

= [1 - cos(x)]/[1 - cos²(x)] = [1 - cos(x)]/sin²(x)

= 1/sin²(x) - cos(x)/sin²(x) = csc²(x) - (cot(x))(csc(x))

Now we can integrate.

∫1/(cos(x) + 1)dx = ∫{csc²(x) - (cot(x))(csc(x))}dx

= -cot(x) + csc(x) + C