some algebraic mistake (I guess) while doing integral

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While I was doing integration. I got an example which looks like,

$$\int \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}}dx$$

I just took $\sin^{-1}x=z$. At the last moment of integration. I got an equation something just like this (Here most of people are very professional that's why I am not writing all steps cause, the question will become very big).

$$\int z^2 dz$$ $$=\frac{z^3}{3}+c$$ $$=\frac{(\sin^{-1}x)^3}{3}+c$$

In my book, they have written. $$\frac{z^3}{3}+c=\frac{1}{3}\cdot (\sin^{-1}x)^2+c$$

I guess power of $\sin^{-1}x$ should cube.

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