2.5 Simplification by Substitution

We see $\sqrt{1+x^2}$, so let's substitute $x = \tan \theta$, where $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) - \{0\}$. Then $\theta = \tan^{-1} x$.

Expression becomes: $\tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right)$ Convert to sine and cosine: $= \tan^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$ Use half-angle formulas: $1 - \cos\theta = 2\sin^2(\theta/2)$ $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ $= \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right)$ $= \tan^{-1}\left(\tan(\frac{\theta}{2})\right)$ Since $\theta/2$ is in the principal branch, this simplifies to: $= \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$