why is the following U-Substitution wrong?

It is known that

$$\int \frac{1}{\sqrt{1-x^2}} dx = arcsin(x)+c$$

this can be done utilizing u-substitution $$x = sin(u)$$

However,

i can let $$u = 1-x^2$$

$$dx = -2u \, du$$

which gives the integral

$$-\int \frac{2u}{\sqrt{u}} du$$

which then returns $$\frac{-4\sqrt{u^3}}{3}$$

which is clearly wrong.

So why is this wrong?

thanks very much for your help and apologies about the elementary question.

REMARK

Upon checking other posts , a common problem has to do with the range of the function being substituted in.

Example

What is wrong with the following u-substitution?

• If $u = 1-x^2$ then $d\color{red}u = -2\color{red}x d\color{red}x$ – J. W. Tanner May 2 at 5:18
• Your differentials are incorrect. With your u-sub $du = -2xdx$. This is not the same as $dx = -2udu$. I don't think that a u-sub is possible here. This is just a well-known integral, which is known based on the fact that the derivative of $\sin^{-1}(x) = 1/(\sqrt{1-x^2})$. – D.B. May 2 at 5:19

Your alternative method is incorrect, because if $$u=1-x^2$$ then $$du=-2x\,dx$$.
You wrote $$dx=-2u\,du$$; that is different.