simplify $\sin^{-1}(\tan x)$

Question

I'm aware of how we can simplify functions which have $Arc$ as an argument . For example $\sin(\cos^{-1}(x)) = \sqrt{1-x^2}$ but what about cases which $Arc$ is out of the parentheses ? For instance consider this : $\sin^{-1}(\tan x)$ . Is there any way for simplification ?

Answer 1

Well, let $f(x)=\arcsin(\tan(x))$, $x\in(-\frac{\pi}{4},\frac{\pi}{4})$. Now, since: $$\arcsin(x)=\int_0^{x}\frac{1}{\sqrt{1-t^2}}dt$$ we have, that: $$f(x)=\int_0^{\tan(x)}\frac{1}{\sqrt{1-t^2}}dt$$ So: $$\begin{align*}f'(x)=&\frac{1}{\sqrt{1-\tan^2(x)}}\frac{1}{\cos^2(x)}=\frac{1}{\cos^2(x)\sqrt{1-\frac{\sin^2(x)}{\cos^2(x)}}}=\\=&\frac{1}{\cos^2(x)\sqrt{\frac{cos^2(x)-\sin^2(x)}{\cos^2(x)}}}=\frac{1}{\cos(x)\sqrt{\cos(2x)}}\end{align*}$$ So: $$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}dt+c$$ and since $f(0)=\arcsin(\tan(0))=0$, we have $c=0$, so: $$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}d,\ x\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$$