Question Number 217755 by Tawa11 last updated on 20/Mar/25
$$\int\:\frac{\mathrm{cos}\left(\mathrm{sin}^{−\:\mathrm{1}} \mathrm{x}\right)\:+\:\mathrm{cos}^{−\:\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}\:+\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}}\right)\right.\right.}\:\mathrm{dx} \\ $$
Commented by mr W last updated on 20/Mar/25
$${you}\:{can}\:{even}\:{make}\:{it}\:{more}\:{nice} \\ $$$${looking} \\ $$$$\int\:\frac{\mathrm{cos}\left(\mathrm{sin}^{−\:\mathrm{1}} \mathrm{x}+{e}^{\mathrm{tan}\:{x}} \right)\:+\:\mathrm{cos}^{−\:\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}+\mathrm{tanh}\:{x}\right)}{\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}\:+\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}}\right)\right.\right.}\:\mathrm{dx} \\ $$$${there}\:{are}\:{a}\:{lot}\:{of}\:{other}\:{mathematic} \\ $$$${functions}\:{and}\:{symbols}\:{which}\:{can} \\ $$$${still}\:{be}\:{put}\:{into}\:{the}\:{formula}. \\ $$
Commented by Tawa11 last updated on 20/Mar/25
$$\mathrm{Hahahahaha}!!! \\ $$
Answered by SdC355 last updated on 20/Mar/25
$$\mathrm{Hint}. \\ $$$$\mathrm{cos}\left(\mathrm{asin}\left({z}\right)\right)=\sqrt{\mathrm{1}−{z}^{\mathrm{2}} } \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\theta\right)+\mathrm{cos}^{\mathrm{2}} \left(\theta\right)=\mathrm{1} \\ $$$$\mathrm{sin}\left(\theta\right)=\pm\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\theta\right)}\:\:\:\mathrm{and}\:\theta=\mathrm{acos}\left({z}\right) \\ $$$$\mathrm{sin}\left(\mathrm{acos}\left({z}\right)\right)=\pm\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{acos}\left({z}\right)\right)}=\sqrt{\mathrm{1}−{z}^{\mathrm{2}} } \\ $$$$\because\:{f}\ast{f}^{−\mathrm{1}} ={z}\:\:\left\{\ast\:\mathrm{is}\:\mathrm{function}\:\mathrm{composition}\:\mathrm{operator}\right\} \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{sin}\left({z}\right)\right)=\frac{\pi}{\mathrm{2}}−{z} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{no}\:\mathrm{one}\:\mathrm{can}\:\mathrm{solve}\:\int\:\:\mathrm{d}{z}\:\:\frac{\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }+\frac{\pi}{\mathrm{2}}−{z}}{\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}+\sqrt{{z}+\sqrt{{z}}}\right)\right)\right)}\:\: \\ $$
Commented by Tawa11 last updated on 20/Mar/25
$$\mathrm{That}\:\mathrm{is}\:\mathrm{serious}\:\mathrm{sir}. \\ $$
Commented by SdC355 last updated on 20/Mar/25
$$\mathrm{Because}\:\:\int\:\:\frac{\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }+\frac{\pi}{\mathrm{2}}−{z}}{''\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}+\sqrt{{z}+\sqrt{{z}}}\right)\right)\right)''} \\ $$
Answered by Marzuk last updated on 20/Mar/25
$${Nah}\:{boy}.{We}\:{might}\:{put}\:{some}\:{function}\:{acrylic} \\ $$$${here}\:{it}\:{is}. \\ $$$$\int\:\frac{\left(\lambda{x}.\Gamma\left({x}\right)+\beta\left({x},{x}+\mathrm{1}\right)+\Theta\left({x}\right)+\alpha\left({x}\right).\Re\left({e}^{{i}\:{cos}^{−\mathrm{1}} {x}} \right)+\:\frac{\partial}{\partial{x}}\:\left[\:\Gamma\left({x}\right)+\beta\left({x},{x}+\mathrm{1}\right)+\Theta\left({x}\right)\right.\right.}{{ln}\left({ln}\left({ln}\left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\right)\right)\right)+{cosh}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\int_{\mathrm{0}} ^{\:{x}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{det}\begin{pmatrix}{{x}\:\:\:\:\:\:{cos}\left({x}\right)}\\{{sin}\left({x}\right)\:\:\:\:{x}^{\mathrm{2}} }\end{pmatrix}\:+\:\psi\left({x}\right)+\Theta\left({x}\right)+\alpha\left({x}\right)}\:{dx} \\ $$$${Now}\:{the}\:{painting}\:{is}\:{perfect} \\ $$