Simplifying an expression involving integrals: $ \xi= (e^\Phi)(\Omega^2)(\Psi^{-1})$ **Simplify the following expression** : $$\Large \xi= (e^\Phi)(\Omega^2)(\Psi^{-1})$$ * * * Where: $$ \large \Phi^{-1}=(e)^{\left[\displaystyle\int\left(\frac{\big(\cos^{-1}\sqrt{1-x^2}\big)^{-1}}{\ln\frac{\big(sin^{-1}\big(2x\sqrt{1-x^2}\big)\big)}{\pi}}\right)dx\right]}$$ _Note: It is given, that the ' $+C$ ' after evaluation of the indefinite integral here, is $0$. (i.e. Assume $C=0$)_ * * * $$\large\Omega= \int^{\infty}_{-\infty}\cos(x^2)+\sin(x^2)dx$$ * * * $$\large \Psi= \displaystyle\int^1_x \frac{1}{\sqrt{\frac{1-z^2}{16}}} dz$$ * * * Note: 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!

$$ \large \Phi^{-1}=(e)^{\left[\displaystyle\int\left(\frac{\big(\cos^{-1}\sqrt{1-x^2}\big)^{-1}}{\ln\frac{\big(sin^{-1}\big(2x\sqrt{1-x^2}\big)\big)}{\pi}}\right)dx\right]}$$ The integral itself evaluates to: $$-\ln\left[\ln\left(\frac{2\arccos(x)}{\pi}\right)\right] + C $$ [Note: Ignore the C as stated in the question (i.e. it's zero)]. Therefore, it becomes $$\large \Phi=\frac{1}{e^{-\ln\left[\ln\left(\frac{2\arccos(x)}{\pi}\right)\right]}}=\ln\left(\frac{2\arccos(x)}{\pi}\right)$$ Therefore $$\color{red} {\large e^\Phi= \frac{2\arccos(x)}{\pi}}$$

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On evaluating, $$\large\Omega=\sqrt{2\pi}$$Therefore $$\color{red}{\large\Omega^2=2\pi}$$

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On evaluating,$$\large \Psi= 4\arccos(x)$$Therefore$$\color{red}{\large \Psi^{-1}= \frac{1}{4\arccos(x)}}$$

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Therefore: $$\color{red}{\Large \xi= (e^\Phi)(\Omega^2)(\Psi^{-1})=1}$$