Simplify the following expression: $$ \xi= (e^\Phi)(\Omega^2)(\Psi^{-1})$$ Where: $$ \log(\Phi^{-1})=\int\frac{(\cos^{-1}\sqrt{1-x^2})^{-1}}{\ln\left(\sin^{-1}(2x\sqrt{1-x^2})/\pi\right)} dx$$
Note: It is given, that the ' $+C$ ' after evaluation of the indefinite integral here, is $0$. (i.e. Assume $C=0$)
$$\Omega= \int^{\infty}_{-\infty}\cos(x^2)+\sin(x^2)dx$$ $$\Psi= \int^1_x \frac{1}{\sqrt{\frac{1-z^2}{16}}} dz$$
I made this question up while solving some integrals, and just wanted to keep a record of it. Hence I've uploaded it in a Q&A style. To whosoever reading this, I hope you find the question mildly interesting!
$$\Phi^{-1}=\exp\int\frac{\big(\cos^{-1}\sqrt{1-x^2}\big)^{-1}}{\ln\frac{\big(\sin^{-1}\big(2x\sqrt{1-x^2}\big)\big)}{\pi}}dx$$ The integral itself evaluates to: $$-\ln\left[\ln\left(\frac{2\arccos(x)}{\pi}\right)\right] + C $$ (Ignored the $C$ as stated in the question).
Therefore, it becomes $$ \Phi=\frac{1}{\exp\left(-\ln\left[\ln\left(\frac{2\arccos(x)}{\pi}\right)\right]\right)}=\ln\left(\frac{2\arccos(x)}{\pi}\right)$$ Therefore $$ \exp(\Phi)= \frac{2\arccos(x)}{\pi}$$
On evaluating, $$\Omega=\sqrt{2\pi}$$Therefore $$\Omega^2=2\pi$$
On evaluating,$$ \Psi= 4\arccos(x)$$Therefore$$ \Psi^{-1}= \frac{1}{4\arccos(x)}$$
Therefore: $$ \xi= (e^\Phi)(\Omega^2)(\Psi^{-1})=1$$