$$ ∫ \, [ \arccos ( x - 1 ) ]^2 \, dx $$
$$ u= \, \arccos ( x - 1 ) $$
$$ \frac {du}{dx}= -\frac1{\sqrt{1-x^2}} $$
$${dx}= - \sqrt{1-x^2}\,\,\, {du} $$
$$ - ∫ \, u^2 \, \sqrt{1-x^2}\,\,\, {du} $$
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$$ \, \arccos ( x - 1 ) =u $$$$ \, ( x - 1 ) =\cos u $$$$ \, x =1+\cos u $$
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$$ - ∫ \, u^2 \, \sqrt{1-(1+\cos u )^2}\,\,\, {du} $$
$$ - ∫ \, u^2 \, \sqrt{1-1+2 \cos u+(\cos u) ^2}\,\,\, {du} $$
$$ - ∫ \, u^2 \, \sqrt{(\cos u) ^2 + 2 \cos u}\,\,\, {du} $$
Hinweis zur partiellen Integration:
$$ \frac{d\, \sqrt{cos^2(u)+2 cos(u)}}{du} = -\frac{ (\sin \, u) (\cos(u)+1)}{\sqrt{(\cos\,u) (\cos\,u+2)} }$$
