Doubt while evaluating $\int_0^{\pi}\frac{1}{\cos x} \text{d}x$

Question

I know that this integral can be evaluated in many ways, but my doubt comes by doing a thing in this specific way.

I was trying to evaluate $$\int_0^{\pi}\frac{1}{\cos x} \text{d}x$$ I would like to multiply for $\frac{\cos x}{\cos x}$, but $\cos x=0$ if $x=\frac{\pi}{2}$ and the integration interval contains $x=\frac{\pi}{2}$; so I suspect that the fractions $\frac{1}{\cos x}$ and $\frac{\cos x}{\cos^2 x}$ aren't equivalent because of this.

So my doubts are:

(1) Can I avoid this problem by using the fact that two integrals are the same if I separate the integration interval in a finite number of points? Like this: $$\int_0^{\pi}\frac{1}{\cos x} \text{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{\cos x} \text{d}x+\int_\frac{\pi}{2}^{\pi}\frac{1}{\cos x} \text{d}x$$ But I'm suspicious, because both integrals on RHS are divergent while the one on LHS isn't. What is the error in doing this? I suspect that this comes by the fact that the fractions aren't equivalent. Am I right? If I'm wrong, what is the reason why this doesn't work?

(2) When am I allowed to manipulate the integrand function? Only when I'm sure that the integrand functions are equivalent in the integration interval?

Thanks for your time.

Answer 1

Note that when we solve this as an improper integral we get a limit that tends to infinity, so we can only assign a Cauchy Principal Value to this integral which is $0$. This is quite logical if you use the graph of $y=\sec x$ to determine the area under the curve from $x=0$ to $x=\pi$. By the symmetry about the asymptote $x=\pi/2$, the algebraic area is $0$ and so is the value of the integral.

$$\begin{aligned}\int_{0}^{\pi}\sec x\mathrm dx &=\lim_{a\to \pi/2}\left(\int_{0}^{a}\sec x\mathrm dx+\int_{a}^{\pi}\sec x\mathrm dx\right)\\ &=\bigg[\ln |\sec x+\tan x|\bigg]_{0}^{\pi/2}+\bigg[\ln |\sec x+\tan x|\bigg]_{\pi/2}^{\pi}\\&=\left(\lim_{x\to \pi/2}\ln|\sec x+\tan x|-0\right)+\left(0-\lim_{x\to \pi/2}\ln|\sec x+\tan x|\right)\end{aligned}$$