I have to solve the following integral: $\int (1-x^2)^{3/2} dx$, $|x|<1$. I did the following substitution:
$z:= 1-x^2 \Rightarrow \frac{dz}{dx}=-2x \Rightarrow dx=\frac{dz}{-2x}$
$\int (1-x^2)^{3/2} dx \stackrel{z=\dots}{=} \int z^{3/2} \frac{dz}{-2x}=-\frac{1}{2x}\cdot\int z^{3/2}dz = -\frac{1}{2x} \cdot \frac{2}{5}z^{5/2} \stackrel{z=...}{=} -\frac{1}{10}x \cdot (1-x^2)^{5/2}$
I thought, my solution was right. Then I checked Wolfram Mathematica for its solution, and it said, the solution was $\int(1-x^2)^{3/2}dx = \frac{1}{8} \left(x\cdot\sqrt{1-x^2}\cdot(5-2x^2)+3\cdot\sin^{-1}(x)\right)$ ($sin^{-1}$ being the inverse $sin$).
I requested the step-by-step-solution, and it said, I had to substitute $x=sin(u) \Rightarrow dx=\cos(u) du$ like this.
Why can't I substitute by $z:= 1-x^2$, as it gives obviously the wrong solution? I understand why substituting with $sin$ makes sense, but I wouldn't get that idea on my own. And I don't see where substituting with $z:=1-x^2$ is wrong. So how can I decide, that this substitution is the wrong function, and I have to look for another? Any help is very appreciated!