# Mathematica outputs a trigonometric integral ($\sec^3$) in a form I can't prove

The indefinite integral is of course $$1/2 ( \sec(x) \tan(x) + \ln | \sec(x) + \tan(x) | ( + C)$$.

Mathematica gives:

Integrate[Sec[x]^3, x]

1/2 (-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x] Tan[x])


The $$1/2 \sec(x) \tan(x)$$ is there, but I've spent a couple of hours trying to prove that Mathematica's logarithm really is $$\ln | \sec(x) + \tan(x) |$$, and I just can't do it! The $$x/2$$ half-angles throw a spanner into the works for me. They just seem so wrong to me, it's like the half-angle formula backwards. I get square roots where I'd like to see squares.

I'm sure I'm missing something obvious, but I just can't see it!

• One thing you must understand about Mathematica is that it assumes all variables are complex-valued, unless informed otherwise. Thus, $\log|\sec x+\tan x|$ is an admissible antiderivative for real $x$, but is not the right answer in general. Aug 16 '20 at 3:49
• I didn't know that, thanks! I really should order Wolfram Mathematica for dummies or something. Aug 16 '20 at 17:03

Differentiate, combine the logarithms, and work backwards using the half angle formulae and the identity $$1+\tan(x)^2 = \sec(x)^2$$

FullSimplify[
D[1/2 (-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x] Tan[x]), x]
]
(* result: Sec[x]^3 *)


You can get there yourself if you first show:

FullSimplify[-(-(1/2) Cos[x/2] - 1/2 Sin[x/2])/(
Cos[x/2] - Sin[x/2]) + (1/2 Cos[x/2] - 1/2 Sin[x/2])/(
Cos[x/2] + Sin[x/2])]

(* Sec[x] *)


To get the above result, take a look at what happens when you put it all over a common denominator:

Together[-((-(1/2) Cos[x/2] - 1/2 Sin[x/2])/(Cos[x/2] - Sin[x/2])) + (
1/2 Cos[x/2] - 1/2 Sin[x/2])/(Cos[x/2] + Sin[x/2])]

(* (Cos[x/2]^2 + Sin[x/2]^2)/
((Cos[x/2] - Sin[x/2]) (Cos[x/2] + Sin[x/2])) *)


The numerator is obviously 1 by the identity $$\cos(\theta)^2+\sin(\theta)^2=1$$ and the denominator is $$\cos(x)$$ by half angles. To see this, expand the denominator $$d=\left(\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right) \left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right)$$ to get $$d=\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)$$. Then we have $$d = 1-2 \sin ^2\left(\frac{x}{2}\right) = \cos(x)$$ and $$1/d$$ is $$\sec(x)$$

... and as for the rest of the derivative:

FullSimplify[1 - Sec[x]^2]
(* Tan[x]^2 *)


So therefore:

D[1/2 (-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x] Tan[x]), x]

(* 1/2 (Sec[x]^3 - (-(1/2) Cos[x/2] - 1/2 Sin[x/2])/(
Cos[x/2] - Sin[x/2]) + (1/2 Cos[x/2] - 1/2 Sin[x/2])/(
Cos[x/2] + Sin[x/2]) + Sec[x] Tan[x]^2) *)

(* == (Sec[x]^3 + Sec[x] (1 + Tan[x]^2))/2 *)
(* == (Sec[x]^3 + Sec[x]^3)/2 == Sec[x]^3 *)

• Thanks! I got as far as (1 + tan x/2) / (1 - tan x/2). That's when I gave up. I think I should have kept grinding, because I just found the identity tan(x/2 + pi/4) = sec(x) + tan(x). Aug 15 '20 at 15:49