Prove that an integral is real and ignore complex parts

Question

(This question comes from this video)

Imagine we are trying to solve the integral :

$$I=\int_{0}^{\pi/2} \ln(2\cos x)\,dx$$ This integral is defined on the real numbers. However, if I decide to use eulers complex cosine formula ($\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$), I can reduce the integral to this (this is shown in the video) :

$$ I = \frac{i\pi^2}{8} - i( 1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{9^2}+\cdots)$$

As you can the the result found this way is a pure imaginary number. In the video, it is then explained that since the integral is defined on the real numbers, therefor the complex result is simply a "residue" of the calculations. My question is, is this reasoning valid ? If not, what could bring us to the same result, that $I = 0$ (I doubt the youtube video would assume something wrong)

I turns out that the result is correct and saying that the result is equal to zero means that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{9^2}+\cdots = \frac{\pi^2}{8}$. We can the continue and use this result to show that $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$.

Answer 1

One can prove the integration result by pure real methods.

First using the substitution $t=\frac\pi2-x$ observe: $$ I=\int_0^{\pi/2}\log(2\cos x)dx=\int_0^{\pi/2}\log(2\sin x)dx. $$

Next add both above integrals to obtain $$\begin{align} 2I&=\int_0^{\pi/2}\log(4\cos x\sin x)dx\tag1\\ &=\int_0^{\pi/2}\log(2\sin 2x)dx\tag2\\ &=2\int_0^{\pi/4}\log(2\sin 2x)dx\tag3\\ &=\int_0^{\pi/2}\log(2\sin t)dt\tag4\\ &=I \end{align}$$ where we used:

1. $\log a+\log b=\log(ab)$
2. $2\cos x\sin x=\sin 2x$
3. $\sin(x)=\sin(\pi-x)$
4. $x\mapsto \frac t2$

Finally: $$2I=I\implies I=0. $$