What is $I=\int\frac{dx}{\sqrt{x^2+1}}$?
Mitsuo Sugiura's answer in his "Introduction to Analysis 1" is like the following:
Let $x=\tan\theta$.
Then, $I=\int\frac{\sec^2\theta}{\sec\theta}d\theta=\int\frac{d\theta}{\cos\theta}=\log\left|\frac{1+\sin\theta}{\cos\theta}\right|=\log\left|x+\sqrt{x^2+1}\right|$.
Sugiura calculated $\sqrt{x^2+1}=\sec\theta$.
But what happens if $\sqrt{x^2+1}=-\sec\theta$?
Let $\varphi:(\frac{\pi}{2},\frac{3\pi}{2})\ni\theta\mapsto\tan\theta\in\mathbb{R}.$
Let $t=\varphi(\theta)$.
Then,
$dt=\phi'(\theta)d\theta=\sec^2\theta d\theta$,
$\sqrt{t^2+1}=\sqrt{(\varphi(\theta))^2+1}=\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=-\sec\theta$.
Let $c$ be a fixed point in $(\frac{\pi}{2},\frac{3\pi}{2})$.
Let $x\in (\frac{\pi}{2},\frac{3\pi}{2})$.
$I=\int_{c}^{x}\frac{dt}{\sqrt{t^2+1}}+C=\int_{\varphi^{-1}(c)}^{\varphi^{-1}(x)}\frac{\sec^2\theta d\theta}{-\sec\theta}+C=\int_{\varphi^{-1}(c)}^{\varphi^{-1}(x)}(-\sec\theta) d\theta+C=-\log\left(-\frac{1+\sin(\varphi^{-1}(x))}{\cos(\varphi^{-1}(x))}\right)+C'=-\log\left(-\tan(\varphi^{-1}(x))-\frac{1}{\cos(\varphi^{-1}(x))}\right)+C'=-\log(-x+\sqrt{x^2+1})+C'=\log(x+\sqrt{x^2+1})+C'$.
Even if $\sqrt{x^2+1}=-\sec\theta$, Sugiura's answer is correct.
Is this a coincidence?