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How do I integrate $\ z_1^{19}\cdot(z_2-z_1)^{19}\cdot(1-z_2)^{19}$ with respect to $z_2$ given $0 < z_1 < z_2 < 1$?
I am unsure what substitution to use.

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If you want to integrate $z_2$ from $z_1$ to $1$ then : $I(z_1)=z_1^{19} \int_{z_1}^1 (z_2-z_1)^{19}\cdot(1-z_2)^{19} dz_2$

Set $t:=\dfrac{z_2-z_1}{1-z_1}$ then $z_2=z_1+t\cdot(1-z_1)$ so that :
$z_2-z_1=t\cdot (1-z_1)$ and $1-z_2=(1-t)\cdot(1-z_1)$

I'll let you conclude,