Find the interval for which $2\arctan x + \arcsin \frac{2x}{1+x^{2}}$ is an independent of x?

Question

I used formula and simplified the expression to $\Rightarrow 4\tan^{-1} x$

Answer 1

HINT:

Let $\arctan x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2\iff-\pi\le2y\le\pi$

$\arcsin\dfrac{2x}{1+x^2}=\arcsin(\sin2y)$ $=\begin{cases} 2y &\mbox{if } -\dfrac\pi2\le2y\le\dfrac\pi2 \\ \pi-2y & \mbox{if } 2y>\dfrac\pi2 \\ -\pi-2y & \mbox{if } 2y<-\dfrac\pi2 \end{cases}$

Answer 2

The function $f(x)=4tan^{-1}(x)$ is invertible on $\mathbb R$.

The inverse function is $g(x)=\tan(\frac{x}{4})$ with $x\in (-2\pi,2\pi)$.

Answer 3

The region of independence is outside the unit circle, i.e. |x|>1. Inside the unit circle the approximation you have used applies.

Answer 4

We may replace $x$ with $\tan\theta$: in order to study the behaviour of $2\arctan x+\arcsin\frac{2x}{1+x^2}$, we may just study the behaviour of $$ g(\theta) = 2\theta + \arcsin(\sin(2\theta)) $$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. $g(\theta)$ equals $4\theta$ over $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$, $-\pi$ over $\left(-\frac{\pi}{2},-\frac{\pi}{4}\right]$ and $\pi$ over $\left[\frac{\pi}{4},\frac{\pi}{2}\right)$, hence the original function is constant over $[1,+\infty)$ and $(-\infty,-1]$.

Here it is the graph of the original function over $[-2,2]$:

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